{"id":120,"date":"2017-12-14T21:26:29","date_gmt":"2017-12-14T21:26:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/types-of-chemical-reactions-single-and-double-displacement-reactions\/"},"modified":"2022-11-14T02:23:38","modified_gmt":"2022-11-14T02:23:38","slug":"types-of-chemical-reactions-single-and-double-displacement-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/types-of-chemical-reactions-single-and-double-displacement-reactions\/","title":{"raw":"7.3 Classifying Chemical Reactions","rendered":"7.3 Classifying Chemical Reactions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define five common types of chemical reactions (single-replacement, double-replacement, composition, decomposition, and combustion).<\/li>\r\n \t<li>Classify chemical reactions as one of these five types given appropriate descriptions or chemical equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<p id=\"ball-ch04_s02_p01\" class=\"para editable block\">Up to now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of how to classify certain types of chemical reactions that allow us to predict what the products of the reaction will be.<\/p>\r\n\r\n<h2>Composition Reactions<\/h2>\r\n<\/div>\r\n<p id=\"ball-ch04_s04_p02\" class=\"para editable block\">A <strong>composition reaction<\/strong> (sometimes also called a <em class=\"emphasis\">combination reaction<\/em> or a <em class=\"emphasis\">synthesis reaction<\/em>) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{2 H}}_{2}\\text{(}g\\text{)}+{\\text{O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{2 H}}_{2}\\text{O(}l\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p03\" class=\"para editable block\">water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance\u2014water\u2014as a product. So this is a composition reaction.<\/p>\r\n\r\n<h2>Decomposition Reactions<\/h2>\r\n<p id=\"ball-ch04_s04_p04\" class=\"para editable block\">A <strong>decomposition reaction<\/strong> starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{2 NaHCO}}_{3}\\text{(}s\\text{)}\\rightarrow{\\text{ Na}}_{2}\\text{CO}_{3}\\text{(}s\\text{)}+{\\text{ CO}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p05\" class=\"para editable block\">sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.<\/p>\r\n<p id=\"ball-ch04_s04_p06\" class=\"para editable block\">Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.<\/p>\r\n\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0 <strong>Composition and Decomposition Reactions<\/strong><\/h3>\r\n<p id=\"ball-ch04_s04_p07\" class=\"para\">Identify each equation as a composition reaction, a decomposition reaction, or neither.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\large{\\text{Fe}}_{2}\\text{O}_{3}\\text{(}s\\text{)}+\\text{3 SO}_{3}\\text{(}g\\text{)}\\rightarrow {\\text{Fe}}_{2}{\\text{(}{\\text{SO}}_{4}\\text{)}}_{3}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{NaCl}}\\text{(}aq\\text{)}+{\\text{ AgNO}}_{3}\\text{(}aq\\text{)}\\rightarrow{\\text{NaNO}}_{3}\\text{(}aq\\text{)}+\\text{AgCl(}s\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{Cr}}_{2}\\text{O}_{7}\\text{(}s\\text{)}\\rightarrow\\text{Cr}_{2}\\text{O}_{3}\\text{(}s\\text{)}+{\\text{4 H}}_{2}{\\text{O}}\\text{(}l\\text{)}+{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"509992\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"509992\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>In this equation, two substances combine to make a single substance. This is a composition reaction.<\/li>\r\n \t<li>Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.<\/li>\r\n \t<li>A single substance reacts to make multiple substances. This is a decomposition reaction.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nIdentify the equation as a composition reaction, a decomposition reaction, or neither.\r\n\r\n[latex]\\large{\\text{C}}_{3}\\text{H}_{8}\\text{(}g\\text{)}\\rightarrow {\\text{C}}_{3}\\text{H}_{4}\\text{(}g\\text{)}+\\text{H}_{2}\\text{(}g\\text{)}[\/latex]\r\n\r\n[reveal-answer q=\"25094\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"25094\"]decomposition reaction[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Single-Replacement Reactions<\/h2>\r\nA <strong>single-replacement reaction<\/strong> (sometimes referred to as a <em>single-displacement reaction<\/em>) is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example,\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 HCl}\\text{(}aq\\text{)}+{\\text{Zn}}\\text{(}s\\text{)}\\rightarrow{\\text{ZnCl}}_{2}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s02_p03\" class=\"para editable block\">is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element\u2014hydrogen\u2014is formed. Another example of a single-replacement reaction is<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 NaCl}\\text{(}aq\\text{)}+{\\text{F}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{2 NaF}}\\text{(}aq\\text{)}+{\\text{Cl}}_{2}\\text{(}g\\text{)}[\/latex]<\/span><\/span><\/p>\r\n\r\n<h2>Double-Replacement Reactions<\/h2>\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<p id=\"ball-ch04_s02_p18\" class=\"para editable block\">A <strong>double-replacement reaction<\/strong> (sometimes referred to as a <em>double-displacement reaction<\/em>) occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{CuCl}}_{2}\\text{(}aq\\text{)}+{\\text{2 AgNO}}_{3}\\text{(}aq\\text{)}\\rightarrow \\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+\\text{2 AgCl(}s\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s02_p19\" class=\"para editable block\">There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.<\/p>\r\n\r\n<\/div>\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0 <strong>Replacement Reactions<\/strong><\/h3>\r\n<p id=\"ball-ch04_s04_p07\" class=\"para\">Predict the products of this double-replacement equation:<\/p>\r\n<p class=\"para\">[latex]\\large{\\text{Na}}_{2}\\text{SO}_{4}+\\text{BaCl}_{2}\\rightarrow{\\text{?}}[\/latex]<\/p>\r\n[reveal-answer q=\"509994\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"509994\"]\r\n\r\nThinking about the reaction as either switching the cations or switching the anions. The barium cation, Ba<sup>2+<\/sup>, will combine with a sulfate anion, SO<sub>4<\/sub><sup>2-<\/sup>, to give the net-neutral ionic compound barium sulfate, BaSO<sub>4<\/sub>. The sodium cation, Na<sup>+<\/sup>, will combine with a chloride anion, Cl<sup>-<\/sup>, to give the net-neutral ionic compound sodium chloride, NaCl.\r\n\r\n[latex]\\large{\\text{Na}}_{2}\\text{SO}_{4}+\\text{BaCl}_{2}\\rightarrow{\\text{2 NaCl}+\\text{Ba}}\\text{SO}_{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nPredict the products of this double-replacement equation:\r\n\r\n[latex]\\large{\\text{KBr}}+{\\text{ AgNO}}_{3}\\rightarrow{\\text{?}}[\/latex]\r\n\r\n[reveal-answer q=\"25095\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"25095\"]KNO<sub class=\"subscript\">3<\/sub> and AgBr[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<p id=\"ball-ch04_s02_p24\" class=\"para editable block\">Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A\u00a0<strong>precipitation reaction<\/strong> occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid\u00a0<strong>precipitate<\/strong>. The formation of a solid precipitate is the driving force that makes the reaction proceed.<\/p>\r\n<p id=\"ball-ch04_s02_p25\" class=\"para editable block\">To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use\u00a0<strong>solubility rules<\/strong>, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble).\u00a0 <a class=\"xref\" href=\"#ball-ch04_s02_t01\">Table 1 \"Some Useful Solubility Rules\"<\/a> lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in <a class=\"xref\" href=\"#ball-ch04_s02_t01\">Table 1 \"Some Useful Solubility Rules\"<\/a>. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.<\/p>\r\n\r\n<div id=\"ball-ch04_s02_t01\" class=\"table block\">\r\n<p class=\"title\"><span class=\"title-prefix\">Table 1<\/span> Some Useful Solubility Rules<\/p>\r\n\r\n<table cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td><strong class=\"emphasis bold\">These compounds generally dissolve in water (are soluble):<\/strong><\/td>\r\n<td><strong class=\"emphasis bold\">Exceptions:<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>All compounds of Li<sup class=\"superscript\">+<\/sup>, Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Rb<sup class=\"superscript\">+<\/sup>, Cs<sup class=\"superscript\">+<\/sup>, and NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/td>\r\n<td>None<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>All compounds of NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> and C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup><\/td>\r\n<td>None<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Compounds of Cl<sup class=\"superscript\">\u2212<\/sup>, Br<sup class=\"superscript\">\u2212<\/sup>, I<sup class=\"superscript\">\u2212<\/sup><\/td>\r\n<td>Ag<sup class=\"superscript\">+<\/sup>, Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, Pb<sup class=\"superscript\">2+<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Compounds of SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2<\/sup><\/td>\r\n<td>Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, Pb<sup class=\"superscript\">2+<\/sup>, Sr<sup class=\"superscript\">2+<\/sup>, Ba<sup class=\"superscript\">2+<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong class=\"emphasis bold\">These compounds generally do not dissolve in water (are insoluble):<\/strong><\/td>\r\n<td><strong class=\"emphasis bold\">Exceptions:<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Compounds of CO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">2\u2212<\/sup> and PO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">3\u2212<\/sup><\/td>\r\n<td>Compounds of Li<sup class=\"superscript\">+<\/sup>, Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Rb<sup class=\"superscript\">+<\/sup>, Cs<sup class=\"superscript\">+<\/sup>, and NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Compounds of OH<sup class=\"superscript\">\u2212<\/sup><\/td>\r\n<td>Compounds of Li<sup class=\"superscript\">+<\/sup>, Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Rb<sup class=\"superscript\">+<\/sup>, Cs<sup class=\"superscript\">+<\/sup>, NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>, Sr<sup class=\"superscript\">2+<\/sup>, and Ba<sup class=\"superscript\">2+<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\nA vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:\r\n<p style=\"text-align: center;\">[latex]\\large2\\text{KI(}aq\\text{)}+\\text{Pb}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{PbI}}_{2}\\text{(}s\\text{)}+2{\\text{KNO}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\r\nThis observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.\r\n\r\nLead iodide is a bright yellow solid that was formerly used as an artist\u2019s pigment known as iodine yellow (Figure 1). The properties of pure PbI<sub>2<\/sub> crystals make them useful for fabrication of X-ray and gamma ray detectors.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211224\/CNX_Chem_04_02_LeadIodide1.jpg\" alt=\"A photograph is shown of a yellow green opaque substance swirled through a clear, colorless liquid in a test tube.\" width=\"300\" height=\"329\" \/> Figure 1. A precipitate of PbI<sub>2<\/sub> forms when solutions containing Pb<sup>2+<\/sup> and I<sup>\u2212<\/sup> are mixed. (credit: Der Kreole\/Wikimedia Commons)[\/caption]\r\n\r\nThe solubility guidelines discussed above may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation\/anion pairing could result in an insoluble compound.\r\n\r\nFor example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag<sup>+<\/sup>, NO<sup>\u2013<\/sup>, Na<sup>+<\/sup>, and F<sup>\u2013<\/sup> ions. Aside from the two ionic compounds originally present in the solutions, AgNO<sub>3<\/sub> and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO<sub>3<\/sub> and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equation:\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{NaF}}\\text{(}aq\\text{)}+{\\text{ AgNO}}_{3}\\text{(}aq\\text{)}\\rightarrow{\\text{AgF}}\\text{(}s\\text{)}+{\\text{ NaNO}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\r\n\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 3: <strong>Predicting Precipitate Reactions<\/strong><\/h3>\r\nWill a double-replacement reaction occur? If so, identify the products.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\large{\\text{KBr}}\\text{(}aq\\text{)}+\\text{Ca}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{?}}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{NaOH}}\\text{(}aq\\text{)}+\\text{Fe}\\text{Cl}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{?}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"509995\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"509995\"]\r\n\r\na. According to the solubility rules, both Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)\u2014namely, CaBr<sub class=\"subscript\">2<\/sub> and KNO<sub class=\"subscript\">3<\/sub>. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.\r\n<p class=\"para\">b. According to the solubility rules, both NaOH and FeCl<sub class=\"subscript\">2<\/sub> are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)<sub class=\"subscript\">2<\/sub>. NaCl is soluble, but, according to the solubility rules, Fe(OH)<sub class=\"subscript\">2<\/sub> is not. Therefore, a reaction would occur, and Fe(OH)<sub class=\"subscript\">2<\/sub>(s) would precipitate out of solution. The balanced chemical equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{2 NaOH}}\\text{(}aq\\text{)}+\\text{Fe}\\text{Cl}_{2}\\text{(}aq\\text{)}\\rightarrow\\text{2 NaCl}\\text{(}aq\\text{)}+\\text{Fe}{\\text{(}{\\text{OH}}\\text{)}}_{2}\\text{(}s\\text{)}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWill a double-replacement equation occur? If so, identify the products.\r\n\r\n[latex]\\large\\text{Sr}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+\\text{K}\\text{Cl}\\text{(}aq\\text{)}\\rightarrow\\text{?}[\/latex]\r\n\r\n[reveal-answer q=\"25096\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"25096\"]No reaction; all possible products are soluble.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Combustion Reactions<\/h2>\r\n<p id=\"ball-ch04_s04_p10\" class=\"para editable block\">A\u00a0<strong>combustion reaction <\/strong>occurs when a reactant combines with oxygen, many times from the atmosphere, to produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N<sub class=\"subscript\">2<\/sub>. Many reactants, called <em class=\"emphasis\">fuels<\/em>, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O. For example, the balanced chemical equation for the combustion of methane, CH<sub class=\"subscript\">4<\/sub>, is as follows:<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{CH}}_{4}\\text{(}g\\text{)}+{\\text{ 2 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}+{\\text{2 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p11\" class=\"para editable block\">Kerosene can be approximated with the formula C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">26<\/sub>, and its combustion equation is<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 C}_{12}\\text{H}_{26}\\text{(}l\\text{)}+{\\text{ 37 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{24 CO}}_{2}\\text{(}g\\text{)}+{\\text{26 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p12\" class=\"para editable block\">Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH, whose combustion equation is<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{C}_{2}\\text{H}_{5}\\text{OH}\\text{(}l\\text{)}+{\\text{ 3 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{2 CO}}_{2}\\text{(}g\\text{)}+{\\text{3 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p13\" class=\"para editable block\">If nitrogen is present in the original fuel, it is converted to N<sub class=\"subscript\">2<\/sub>, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>, we have<\/p>\r\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 C}_{2}\\text{H}_{2}\\text{N}_{2}\\text{O}_{4}\\text{(}l\\text{)}+{\\text{ O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{4 CO}}_{2}\\text{(}g\\text{)}+{\\text{ 2 H}}_{2}\\text{O(}g\\text{)}+\\text{ N}_{2}\\text{(}g\\text{)}[\/latex]<\/span><\/span><\/p>\r\n\r\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\r\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 4: <strong>Combustion Reactions<\/strong><\/h3>\r\nComplete and balance each combustion equation.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The combustion of propane, C<sub>3<\/sub>H<sub>8<\/sub><\/li>\r\n \t<li>The combustion of NH<sub>3<\/sub><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"509998\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"509998\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"para\">The products of the reaction are CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O, so our unbalanced equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{C}_{3}\\text{H}_{8}+{\\text{ O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span>\r\n<p id=\"ball-ch04_s04_p15\" class=\"para\">Balancing (and you may have to go back and forth a few times to balance this), we get<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{C}_{3}\\text{H}_{8}+{\\text{ 5 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{ 3 CO}}_{2}\\text{(}g\\text{)}+{\\text{ 4 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/li>\r\n \t<li>The nitrogen atoms in ammonia will react to make N<sub>2<\/sub>, while the hydrogen atoms will react with O<sub>2<\/sub> to make H<sub>2<\/sub>O.<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{N}\\text{H}_{3}\\text{(}g\\text{)}+{\\text{ O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{N}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span>\r\n<p id=\"ball-ch04_s04_p15\" class=\"para\">Balancing (and you may have to go back and forth a few times to balance this), we get<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{4 N}\\text{H}_{3}\\text{(}g\\text{)}+{\\text{ 3 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{ 2 N}}_{2}\\text{(}g\\text{)}+{\\text{ 6 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nComplete and balance the combustion equation for cyclopropanol, C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">6<\/sub>O.\r\n\r\n[reveal-answer q=\"25098\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"25098\"]\r\n\r\n[latex]\\large\\text{C}_{3}\\text{H}_{6}\\text{O}\\text{(}l\\text{)}+{\\text{ 4 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{3 CO}}_{2}\\text{(}g\\text{)}+{\\text{ 3 H}}_{2}\\text{O(}g\\text{)}[\/latex]\r\n\r\n[caption id=\"attachment_3213\" align=\"alignnone\" width=\"400\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/5345065044_0d15179564_b.jpg\"><img class=\"wp-image-3213\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212643\/5345065044_0d15179564_b-1.jpg\" alt=\"Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. Source: \u201cflowers and propane\u201d by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic\" width=\"400\" height=\"408\" \/><\/a> Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. Source: \u201cflowers and propane\u201d by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"ball-ch04_s02_n07\" class=\"key_takeaways editable block\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Takeaways<\/h3>\r\n<ul id=\"ball-ch04_s04_l06\" class=\"itemizedlist\">\r\n \t<li>A composition reaction produces a single substance from multiple reactants.<\/li>\r\n \t<li>A decomposition reaction produces multiple products from a single reactant.<\/li>\r\n \t<li>Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make N<sub class=\"subscript\">2<\/sub>).<\/li>\r\n \t<li>A single-replacement reaction replaces one element for another in a compound.<\/li>\r\n \t<li>A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.<\/li>\r\n \t<li>A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.<\/li>\r\n \t<li>Solubility rules are used to predict whether some double-replacement reactions will occur.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"ball-ch04_s02_qs01_ans\" class=\"qandaset block\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div class=\"question\">\r\n\r\n1. Which is a composition reaction and which is not?\r\n\r\n<\/div>\r\n<p style=\"padding-left: 30px;\">a. NaCl +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0AgCl +\u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 CaO +\u00a0CO<sub class=\"subscript\">2<\/sub> \u2192\u00a0CaCO<sub class=\"subscript\">3<\/sub><\/p>\r\n\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p2\" class=\"para\">2. \u00a0Which is a composition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a. H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 HCl<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 2 HBr +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 HCl +\u00a0Br<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p3\" class=\"para\">3. \u00a0Which is a composition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 2 SO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 SO<sub class=\"subscript\">3<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 6 C +\u00a03 H<sub class=\"subscript\">2<\/sub> \u2192\u00a0C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">6<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p4\" class=\"para\">4. \u00a0Which is a composition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 4 Na +\u00a02 C +\u00a03 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>O +\u00a0CO<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p5\" class=\"para\">5. \u00a0Which is a decomposition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 HCl +\u00a0NaOH \u2192\u00a0NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 CaCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0CaO +\u00a0CO<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p6\" class=\"para\">6. \u00a0Which is a decomposition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 3 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 O<sub class=\"subscript\">3<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 2 KClO<sub class=\"subscript\">3<\/sub> \u2192\u00a02 KCl +\u00a03 O<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p7\" class=\"para\">7. \u00a0Which is a decomposition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Na<sub class=\"subscript\">2<\/sub>O +\u00a0CO<sub class=\"subscript\">2<\/sub> \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub> \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O +\u00a0SO<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p8\" class=\"para\">8. \u00a0Which is a decomposition reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 2 C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">5<\/sub>N<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">6<\/sub> \u2192\u00a03 N<sub class=\"subscript\">2<\/sub> +\u00a05 H<sub class=\"subscript\">2<\/sub>O +\u00a07 CO +\u00a07 C<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">12<\/sub>O<sub class=\"subscript\">6<\/sub> +\u00a06 O<sub class=\"subscript\">2<\/sub> \u2192\u00a06 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p9\" class=\"para\">9. \u00a0Which is a combustion reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">12<\/sub>O<sub class=\"subscript\">6<\/sub> +\u00a06 O<sub class=\"subscript\">2<\/sub> \u2192\u00a06 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 2 Fe<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p10\" class=\"para\">10. \u00a0Which is a combustion reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 CH<sub class=\"subscript\">4<\/sub> +\u00a02 F<sub class=\"subscript\">2<\/sub> \u2192\u00a0CF<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p11\" class=\"para\">11. \u00a0Which is a combustion reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 P<sub class=\"subscript\">4<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 P<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">5<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 2 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p12\" class=\"para\">12. \u00a0Which is a combustion reaction and which is not?<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">2<\/sub><\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub>Cl<sub class=\"subscript\">2<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p13\" class=\"para\">13. \u00a0Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p15\" class=\"para\">14. \u00a0Is it possible for a decomposition reaction to also be a combustion reaction? Give an example to support your case.<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s04_qs01_p17\" class=\"para\">15. \u00a0Complete and balance each combustion equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n16. \u00a0Complete and balance each combustion equation.\r\n<p style=\"padding-left: 30px;\">a.\u00a0 B<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0? (The oxide of boron formed is B<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>.)<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0? (The oxide of sulfur formed is SO<sub class=\"subscript\">2<\/sub>.)<\/p>\r\n<p style=\"padding-left: 30px;\">c.\u00a0 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0? (The oxide of sulfur formed is SO<sub class=\"subscript\">3<\/sub>.)<\/p>\r\n\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p5\" class=\"para\">17. Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\r\n<p class=\"para\" style=\"padding-left: 30px;\">a.\u00a0 Zn +\u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 F<sub class=\"subscript\">2<\/sub> +\u00a0FeI<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\r\n18. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<p style=\"padding-left: 30px;\">a. Li +\u00a0MgSO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 NaBr +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n19. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Sn +\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 Al +\u00a0NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p8\" class=\"para\">20. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Mg +\u00a0HCl \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 HI +\u00a0Br<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p13\" class=\"para\">21. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0NaOH \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p14\" class=\"para\">22. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Ca(C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0HNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> +\u00a0Sr(NO<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p15\" class=\"para\">23. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 K<sub class=\"subscript\">2<\/sub>O +\u00a0MgCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p16\" class=\"para\">24. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Sn(OH)<sub class=\"subscript\">2<\/sub> +\u00a0FeBr<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 CsNO<sub class=\"subscript\">3<\/sub> +\u00a0KCl \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p18\" class=\"para\">25. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> +\u00a0Sr(NO<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s02_qs01_p20\" class=\"para\">26. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 KC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> +\u00a0Li<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 KOH +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\r\n<p id=\"ball-ch04_s02_qs01_p19\" class=\"para\">27. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\r\n<p style=\"padding-left: 30px;\">a.\u00a0 K<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a0SrCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n<p style=\"padding-left: 30px;\">b.\u00a0 NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"25099\"]Show Answers to Select Questions[\/reveal-answer]\r\n[hidden-answer a=\"25099\"]\r\n\r\n<strong>1.\u00a0<\/strong> a) \u00a0not composition, b) \u00a0composition\r\n\r\n<strong>3.\u00a0 <\/strong>a) \u00a0composition, b) \u00a0composition\r\n<div class=\"answer\">\r\n\r\n<strong>5.\u00a0 <\/strong>a) \u00a0not decomposition, b) \u00a0decomposition\r\n<div class=\"answer\">\r\n\r\n<strong>7.\u00a0 <\/strong>a) \u00a0not decomposition, b) \u00a0decomposition\r\n<div class=\"answer\">\r\n\r\n<strong>9.\u00a0 <\/strong>a) \u00a0combustion, b) \u00a0combustion\r\n<div class=\"answer\">\r\n\r\n<strong>11. <\/strong>a) \u00a0combustion, b) \u00a0combustion\r\n<div class=\"answer\">\r\n\r\n<strong>13. <\/strong>Yes; 2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O (answers will vary)\r\n\r\n<strong>15. <\/strong>a) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a06 O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a05 H<sub class=\"subscript\">2<\/sub>O,\u00a0\u00a0 b) \u00a04 CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a03 O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O +\u00a02 N<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<div>\r\n\r\n<strong>17. <\/strong>a) \u00a0Zn +\u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0Fe,\u00a0\u00a0\u00a0 b) \u00a03 F<sub class=\"subscript\">2<\/sub> +\u00a02 FeI<sub class=\"subscript\">3<\/sub> \u2192\u00a03 I<sub class=\"subscript\">2<\/sub> +\u00a02 FeF<sub class=\"subscript\">3<\/sub>\r\n\r\n<strong>19. <\/strong>a) \u00a0Sn +\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0SnSO<sub class=\"subscript\">4<\/sub> +\u00a0H<sub class=\"subscript\">2,\u00a0\u00a0\u00a0 <\/sub>b) \u00a02 Al +\u00a03 NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a02 AlBr<sub class=\"subscript\">3<\/sub> +\u00a03 Ni\r\n\r\n<strong>21. <\/strong>a) \u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 NaOH \u2192\u00a0Zn(OH)<sub class=\"subscript\">2<\/sub> +\u00a02 NaNO<sub class=\"subscript\">3,\u00a0\u00a0\u00a0 <\/sub>b) \u00a02 HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192\u00a02 NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>S\r\n\r\n<strong>23. <\/strong>a) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 KBr \u2192\u00a0PbBr<sub class=\"subscript\">2<\/sub> +\u00a02 KNO<sub class=\"subscript\">3<\/sub><sub class=\"subscript\">,\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sub>b) \u00a0K<sub class=\"subscript\">2<\/sub>O +\u00a0MgCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0K<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> +\u00a0MgO\r\n\r\n<strong>27. <\/strong>a) \u00a02 K<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a03 SrCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0Sr<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>(s) +\u00a06 KCl,\u00a0\u00a0\u00a0 b) \u00a02 NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NaCl +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>(s)\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<p class=\"para\"><\/p>\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define five common types of chemical reactions (single-replacement, double-replacement, composition, decomposition, and combustion).<\/li>\n<li>Classify chemical reactions as one of these five types given appropriate descriptions or chemical equations.<\/li>\n<\/ul>\n<\/div>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<p id=\"ball-ch04_s02_p01\" class=\"para editable block\">Up to now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of how to classify certain types of chemical reactions that allow us to predict what the products of the reaction will be.<\/p>\n<h2>Composition Reactions<\/h2>\n<\/div>\n<p id=\"ball-ch04_s04_p02\" class=\"para editable block\">A <strong>composition reaction<\/strong> (sometimes also called a <em class=\"emphasis\">combination reaction<\/em> or a <em class=\"emphasis\">synthesis reaction<\/em>) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{2 H}}_{2}\\text{(}g\\text{)}+{\\text{O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{2 H}}_{2}\\text{O(}l\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p03\" class=\"para editable block\">water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance\u2014water\u2014as a product. So this is a composition reaction.<\/p>\n<h2>Decomposition Reactions<\/h2>\n<p id=\"ball-ch04_s04_p04\" class=\"para editable block\">A <strong>decomposition reaction<\/strong> starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{2 NaHCO}}_{3}\\text{(}s\\text{)}\\rightarrow{\\text{ Na}}_{2}\\text{CO}_{3}\\text{(}s\\text{)}+{\\text{ CO}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p05\" class=\"para editable block\">sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.<\/p>\n<p id=\"ball-ch04_s04_p06\" class=\"para editable block\">Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.<\/p>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0 <strong>Composition and Decomposition Reactions<\/strong><\/h3>\n<p id=\"ball-ch04_s04_p07\" class=\"para\">Identify each equation as a composition reaction, a decomposition reaction, or neither.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\large{\\text{Fe}}_{2}\\text{O}_{3}\\text{(}s\\text{)}+\\text{3 SO}_{3}\\text{(}g\\text{)}\\rightarrow {\\text{Fe}}_{2}{\\text{(}{\\text{SO}}_{4}\\text{)}}_{3}[\/latex]<\/li>\n<li>[latex]\\large{\\text{NaCl}}\\text{(}aq\\text{)}+{\\text{ AgNO}}_{3}\\text{(}aq\\text{)}\\rightarrow{\\text{NaNO}}_{3}\\text{(}aq\\text{)}+\\text{AgCl(}s\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{Cr}}_{2}\\text{O}_{7}\\text{(}s\\text{)}\\rightarrow\\text{Cr}_{2}\\text{O}_{3}\\text{(}s\\text{)}+{\\text{4 H}}_{2}{\\text{O}}\\text{(}l\\text{)}+{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q509992\">Show Answer<\/span><\/p>\n<div id=\"q509992\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>In this equation, two substances combine to make a single substance. This is a composition reaction.<\/li>\n<li>Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.<\/li>\n<li>A single substance reacts to make multiple substances. This is a decomposition reaction.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Identify the equation as a composition reaction, a decomposition reaction, or neither.<\/p>\n<p>[latex]\\large{\\text{C}}_{3}\\text{H}_{8}\\text{(}g\\text{)}\\rightarrow {\\text{C}}_{3}\\text{H}_{4}\\text{(}g\\text{)}+\\text{H}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25094\">Show Answer<\/span><\/p>\n<div id=\"q25094\" class=\"hidden-answer\" style=\"display: none\">decomposition reaction<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Single-Replacement Reactions<\/h2>\n<p>A <strong>single-replacement reaction<\/strong> (sometimes referred to as a <em>single-displacement reaction<\/em>) is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example,<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 HCl}\\text{(}aq\\text{)}+{\\text{Zn}}\\text{(}s\\text{)}\\rightarrow{\\text{ZnCl}}_{2}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s02_p03\" class=\"para editable block\">is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element\u2014hydrogen\u2014is formed. Another example of a single-replacement reaction is<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 NaCl}\\text{(}aq\\text{)}+{\\text{F}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{2 NaF}}\\text{(}aq\\text{)}+{\\text{Cl}}_{2}\\text{(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<h2>Double-Replacement Reactions<\/h2>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<p id=\"ball-ch04_s02_p18\" class=\"para editable block\">A <strong>double-replacement reaction<\/strong> (sometimes referred to as a <em>double-displacement reaction<\/em>) occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{CuCl}}_{2}\\text{(}aq\\text{)}+{\\text{2 AgNO}}_{3}\\text{(}aq\\text{)}\\rightarrow \\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+\\text{2 AgCl(}s\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s02_p19\" class=\"para editable block\">There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.<\/p>\n<\/div>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0 <strong>Replacement Reactions<\/strong><\/h3>\n<p id=\"ball-ch04_s04_p07\" class=\"para\">Predict the products of this double-replacement equation:<\/p>\n<p class=\"para\">[latex]\\large{\\text{Na}}_{2}\\text{SO}_{4}+\\text{BaCl}_{2}\\rightarrow{\\text{?}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q509994\">Show Answer<\/span><\/p>\n<div id=\"q509994\" class=\"hidden-answer\" style=\"display: none\">\n<p>Thinking about the reaction as either switching the cations or switching the anions. The barium cation, Ba<sup>2+<\/sup>, will combine with a sulfate anion, SO<sub>4<\/sub><sup>2-<\/sup>, to give the net-neutral ionic compound barium sulfate, BaSO<sub>4<\/sub>. The sodium cation, Na<sup>+<\/sup>, will combine with a chloride anion, Cl<sup>&#8211;<\/sup>, to give the net-neutral ionic compound sodium chloride, NaCl.<\/p>\n<p>[latex]\\large{\\text{Na}}_{2}\\text{SO}_{4}+\\text{BaCl}_{2}\\rightarrow{\\text{2 NaCl}+\\text{Ba}}\\text{SO}_{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Predict the products of this double-replacement equation:<\/p>\n<p>[latex]\\large{\\text{KBr}}+{\\text{ AgNO}}_{3}\\rightarrow{\\text{?}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25095\">Show Answer<\/span><\/p>\n<div id=\"q25095\" class=\"hidden-answer\" style=\"display: none\">KNO<sub class=\"subscript\">3<\/sub> and AgBr<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<p id=\"ball-ch04_s02_p24\" class=\"para editable block\">Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A\u00a0<strong>precipitation reaction<\/strong> occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid\u00a0<strong>precipitate<\/strong>. The formation of a solid precipitate is the driving force that makes the reaction proceed.<\/p>\n<p id=\"ball-ch04_s02_p25\" class=\"para editable block\">To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use\u00a0<strong>solubility rules<\/strong>, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble).\u00a0 <a class=\"xref\" href=\"#ball-ch04_s02_t01\">Table 1 &#8220;Some Useful Solubility Rules&#8221;<\/a> lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in <a class=\"xref\" href=\"#ball-ch04_s02_t01\">Table 1 &#8220;Some Useful Solubility Rules&#8221;<\/a>. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.<\/p>\n<div id=\"ball-ch04_s02_t01\" class=\"table block\">\n<p class=\"title\"><span class=\"title-prefix\">Table 1<\/span> Some Useful Solubility Rules<\/p>\n<table cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong class=\"emphasis bold\">These compounds generally dissolve in water (are soluble):<\/strong><\/td>\n<td><strong class=\"emphasis bold\">Exceptions:<\/strong><\/td>\n<\/tr>\n<tr>\n<td>All compounds of Li<sup class=\"superscript\">+<\/sup>, Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Rb<sup class=\"superscript\">+<\/sup>, Cs<sup class=\"superscript\">+<\/sup>, and NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/td>\n<td>None<\/td>\n<\/tr>\n<tr>\n<td>All compounds of NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> and C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup><\/td>\n<td>None<\/td>\n<\/tr>\n<tr>\n<td>Compounds of Cl<sup class=\"superscript\">\u2212<\/sup>, Br<sup class=\"superscript\">\u2212<\/sup>, I<sup class=\"superscript\">\u2212<\/sup><\/td>\n<td>Ag<sup class=\"superscript\">+<\/sup>, Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, Pb<sup class=\"superscript\">2+<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Compounds of SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2<\/sup><\/td>\n<td>Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, Pb<sup class=\"superscript\">2+<\/sup>, Sr<sup class=\"superscript\">2+<\/sup>, Ba<sup class=\"superscript\">2+<\/sup><\/td>\n<\/tr>\n<tr>\n<td><strong class=\"emphasis bold\">These compounds generally do not dissolve in water (are insoluble):<\/strong><\/td>\n<td><strong class=\"emphasis bold\">Exceptions:<\/strong><\/td>\n<\/tr>\n<tr>\n<td>Compounds of CO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">2\u2212<\/sup> and PO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">3\u2212<\/sup><\/td>\n<td>Compounds of Li<sup class=\"superscript\">+<\/sup>, Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Rb<sup class=\"superscript\">+<\/sup>, Cs<sup class=\"superscript\">+<\/sup>, and NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Compounds of OH<sup class=\"superscript\">\u2212<\/sup><\/td>\n<td>Compounds of Li<sup class=\"superscript\">+<\/sup>, Na<sup class=\"superscript\">+<\/sup>, K<sup class=\"superscript\">+<\/sup>, Rb<sup class=\"superscript\">+<\/sup>, Cs<sup class=\"superscript\">+<\/sup>, NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>, Sr<sup class=\"superscript\">2+<\/sup>, and Ba<sup class=\"superscript\">2+<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<p>A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:<\/p>\n<p style=\"text-align: center;\">[latex]\\large2\\text{KI(}aq\\text{)}+\\text{Pb}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{PbI}}_{2}\\text{(}s\\text{)}+2{\\text{KNO}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\n<p>This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.<\/p>\n<p>Lead iodide is a bright yellow solid that was formerly used as an artist\u2019s pigment known as iodine yellow (Figure 1). The properties of pure PbI<sub>2<\/sub> crystals make them useful for fabrication of X-ray and gamma ray detectors.<\/p>\n<div style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211224\/CNX_Chem_04_02_LeadIodide1.jpg\" alt=\"A photograph is shown of a yellow green opaque substance swirled through a clear, colorless liquid in a test tube.\" width=\"300\" height=\"329\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. A precipitate of PbI<sub>2<\/sub> forms when solutions containing Pb<sup>2+<\/sup> and I<sup>\u2212<\/sup> are mixed. (credit: Der Kreole\/Wikimedia Commons)<\/p>\n<\/div>\n<p>The solubility guidelines discussed above may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation\/anion pairing could result in an insoluble compound.<\/p>\n<p>For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag<sup>+<\/sup>, NO<sup>\u2013<\/sup>, Na<sup>+<\/sup>, and F<sup>\u2013<\/sup> ions. Aside from the two ionic compounds originally present in the solutions, AgNO<sub>3<\/sub> and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO<sub>3<\/sub> and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{NaF}}\\text{(}aq\\text{)}+{\\text{ AgNO}}_{3}\\text{(}aq\\text{)}\\rightarrow{\\text{AgF}}\\text{(}s\\text{)}+{\\text{ NaNO}}_{3}\\text{(}aq\\text{)}[\/latex]<\/p>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\n<div class=\"textbox examples\">\n<h3>Example 3: <strong>Predicting Precipitate Reactions<\/strong><\/h3>\n<p>Will a double-replacement reaction occur? If so, identify the products.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\large{\\text{KBr}}\\text{(}aq\\text{)}+\\text{Ca}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{?}}[\/latex]<\/li>\n<li>[latex]\\large{\\text{NaOH}}\\text{(}aq\\text{)}+\\text{Fe}\\text{Cl}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{?}}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q509995\">Show Answer<\/span><\/p>\n<div id=\"q509995\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. According to the solubility rules, both Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)\u2014namely, CaBr<sub class=\"subscript\">2<\/sub> and KNO<sub class=\"subscript\">3<\/sub>. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.<\/p>\n<p class=\"para\">b. According to the solubility rules, both NaOH and FeCl<sub class=\"subscript\">2<\/sub> are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)<sub class=\"subscript\">2<\/sub>. NaCl is soluble, but, according to the solubility rules, Fe(OH)<sub class=\"subscript\">2<\/sub> is not. Therefore, a reaction would occur, and Fe(OH)<sub class=\"subscript\">2<\/sub>(s) would precipitate out of solution. The balanced chemical equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{2 NaOH}}\\text{(}aq\\text{)}+\\text{Fe}\\text{Cl}_{2}\\text{(}aq\\text{)}\\rightarrow\\text{2 NaCl}\\text{(}aq\\text{)}+\\text{Fe}{\\text{(}{\\text{OH}}\\text{)}}_{2}\\text{(}s\\text{)}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Will a double-replacement equation occur? If so, identify the products.<\/p>\n<p>[latex]\\large\\text{Sr}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+\\text{K}\\text{Cl}\\text{(}aq\\text{)}\\rightarrow\\text{?}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25096\">Show Answer<\/span><\/p>\n<div id=\"q25096\" class=\"hidden-answer\" style=\"display: none\">No reaction; all possible products are soluble.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Combustion Reactions<\/h2>\n<p id=\"ball-ch04_s04_p10\" class=\"para editable block\">A\u00a0<strong>combustion reaction <\/strong>occurs when a reactant combines with oxygen, many times from the atmosphere, to produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N<sub class=\"subscript\">2<\/sub>. Many reactants, called <em class=\"emphasis\">fuels<\/em>, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O. For example, the balanced chemical equation for the combustion of methane, CH<sub class=\"subscript\">4<\/sub>, is as follows:<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large{\\text{CH}}_{4}\\text{(}g\\text{)}+{\\text{ 2 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}+{\\text{2 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p11\" class=\"para editable block\">Kerosene can be approximated with the formula C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">26<\/sub>, and its combustion equation is<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 C}_{12}\\text{H}_{26}\\text{(}l\\text{)}+{\\text{ 37 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{24 CO}}_{2}\\text{(}g\\text{)}+{\\text{26 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p12\" class=\"para editable block\">Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH, whose combustion equation is<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{C}_{2}\\text{H}_{5}\\text{OH}\\text{(}l\\text{)}+{\\text{ 3 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{2 CO}}_{2}\\text{(}g\\text{)}+{\\text{3 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p13\" class=\"para editable block\">If nitrogen is present in the original fuel, it is converted to N<sub class=\"subscript\">2<\/sub>, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>, we have<\/p>\n<p style=\"text-align: center;\"><span class=\"informalequation block\"><span class=\"mathphrase\">[latex]\\large\\text{2 C}_{2}\\text{H}_{2}\\text{N}_{2}\\text{O}_{4}\\text{(}l\\text{)}+{\\text{ O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{4 CO}}_{2}\\text{(}g\\text{)}+{\\text{ 2 H}}_{2}\\text{O(}g\\text{)}+\\text{ N}_{2}\\text{(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<div id=\"ball-ch04_s02\" class=\"section\" lang=\"en\">\n<div id=\"ball-ch04_s02_f01\" class=\"figure large editable block\">\n<div class=\"textbox examples\">\n<h3>Example 4: <strong>Combustion Reactions<\/strong><\/h3>\n<p>Complete and balance each combustion equation.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The combustion of propane, C<sub>3<\/sub>H<sub>8<\/sub><\/li>\n<li>The combustion of NH<sub>3<\/sub><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q509998\">Show Answer<\/span><\/p>\n<div id=\"q509998\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"para\">The products of the reaction are CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O, so our unbalanced equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{C}_{3}\\text{H}_{8}+{\\text{ O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p15\" class=\"para\">Balancing (and you may have to go back and forth a few times to balance this), we get<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{C}_{3}\\text{H}_{8}+{\\text{ 5 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{ 3 CO}}_{2}\\text{(}g\\text{)}+{\\text{ 4 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/li>\n<li>The nitrogen atoms in ammonia will react to make N<sub>2<\/sub>, while the hydrogen atoms will react with O<sub>2<\/sub> to make H<sub>2<\/sub>O.<span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{N}\\text{H}_{3}\\text{(}g\\text{)}+{\\text{ O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{N}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span>\n<p id=\"ball-ch04_s04_p15\" class=\"para\">Balancing (and you may have to go back and forth a few times to balance this), we get<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[latex]\\large\\text{4 N}\\text{H}_{3}\\text{(}g\\text{)}+{\\text{ 3 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{ 2 N}}_{2}\\text{(}g\\text{)}+{\\text{ 6 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/span><\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Complete and balance the combustion equation for cyclopropanol, C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">6<\/sub>O.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25098\">Show Answer<\/span><\/p>\n<div id=\"q25098\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\large\\text{C}_{3}\\text{H}_{6}\\text{O}\\text{(}l\\text{)}+{\\text{ 4 O}}_2\\text{(}g\\text{)}\\rightarrow{\\text{3 CO}}_{2}\\text{(}g\\text{)}+{\\text{ 3 H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/p>\n<div id=\"attachment_3213\" style=\"width: 410px\" class=\"wp-caption alignnone\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/5345065044_0d15179564_b.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3213\" class=\"wp-image-3213\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212643\/5345065044_0d15179564_b-1.jpg\" alt=\"Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. Source: \u201cflowers and propane\u201d by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic\" width=\"400\" height=\"408\" \/><\/a><\/p>\n<p id=\"caption-attachment-3213\" class=\"wp-caption-text\">Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. Source: \u201cflowers and propane\u201d by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"ball-ch04_s02_n07\" class=\"key_takeaways editable block\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch04_s04_l06\" class=\"itemizedlist\">\n<li>A composition reaction produces a single substance from multiple reactants.<\/li>\n<li>A decomposition reaction produces multiple products from a single reactant.<\/li>\n<li>Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make N<sub class=\"subscript\">2<\/sub>).<\/li>\n<li>A single-replacement reaction replaces one element for another in a compound.<\/li>\n<li>A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.<\/li>\n<li>A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.<\/li>\n<li>Solubility rules are used to predict whether some double-replacement reactions will occur.<\/li>\n<\/ul>\n<\/div>\n<div id=\"ball-ch04_s02_qs01_ans\" class=\"qandaset block\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"question\">\n<p>1. Which is a composition reaction and which is not?<\/p>\n<\/div>\n<p style=\"padding-left: 30px;\">a. NaCl +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0AgCl +\u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 CaO +\u00a0CO<sub class=\"subscript\">2<\/sub> \u2192\u00a0CaCO<sub class=\"subscript\">3<\/sub><\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p2\" class=\"para\">2. \u00a0Which is a composition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a. H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 HCl<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 2 HBr +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 HCl +\u00a0Br<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p3\" class=\"para\">3. \u00a0Which is a composition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 2 SO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 SO<sub class=\"subscript\">3<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 6 C +\u00a03 H<sub class=\"subscript\">2<\/sub> \u2192\u00a0C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">6<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p4\" class=\"para\">4. \u00a0Which is a composition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 4 Na +\u00a02 C +\u00a03 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>O +\u00a0CO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p5\" class=\"para\">5. \u00a0Which is a decomposition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 HCl +\u00a0NaOH \u2192\u00a0NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 CaCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0CaO +\u00a0CO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p6\" class=\"para\">6. \u00a0Which is a decomposition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 3 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 O<sub class=\"subscript\">3<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 2 KClO<sub class=\"subscript\">3<\/sub> \u2192\u00a02 KCl +\u00a03 O<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p7\" class=\"para\">7. \u00a0Which is a decomposition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Na<sub class=\"subscript\">2<\/sub>O +\u00a0CO<sub class=\"subscript\">2<\/sub> \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub> \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O +\u00a0SO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p8\" class=\"para\">8. \u00a0Which is a decomposition reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 2 C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">5<\/sub>N<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">6<\/sub> \u2192\u00a03 N<sub class=\"subscript\">2<\/sub> +\u00a05 H<sub class=\"subscript\">2<\/sub>O +\u00a07 CO +\u00a07 C<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">12<\/sub>O<sub class=\"subscript\">6<\/sub> +\u00a06 O<sub class=\"subscript\">2<\/sub> \u2192\u00a06 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p9\" class=\"para\">9. \u00a0Which is a combustion reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">12<\/sub>O<sub class=\"subscript\">6<\/sub> +\u00a06 O<sub class=\"subscript\">2<\/sub> \u2192\u00a06 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 2 Fe<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p10\" class=\"para\">10. \u00a0Which is a combustion reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 CH<sub class=\"subscript\">4<\/sub> +\u00a02 F<sub class=\"subscript\">2<\/sub> \u2192\u00a0CF<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p11\" class=\"para\">11. \u00a0Which is a combustion reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 P<sub class=\"subscript\">4<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 P<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">5<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 2 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p12\" class=\"para\">12. \u00a0Which is a combustion reaction and which is not?<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">2<\/sub><\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub>Cl<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p13\" class=\"para\">13. \u00a0Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p15\" class=\"para\">14. \u00a0Is it possible for a decomposition reaction to also be a combustion reaction? Give an example to support your case.<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s04_qs01_p17\" class=\"para\">15. \u00a0Complete and balance each combustion equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p>16. \u00a0Complete and balance each combustion equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 B<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0? (The oxide of boron formed is B<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>.)<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0? (The oxide of sulfur formed is SO<sub class=\"subscript\">2<\/sub>.)<\/p>\n<p style=\"padding-left: 30px;\">c.\u00a0 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0? (The oxide of sulfur formed is SO<sub class=\"subscript\">3<\/sub>.)<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p5\" class=\"para\">17. Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p class=\"para\" style=\"padding-left: 30px;\">a.\u00a0 Zn +\u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p style=\"padding-left: 30px;\">b.\u00a0 F<sub class=\"subscript\">2<\/sub> +\u00a0FeI<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p>18. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a. Li +\u00a0MgSO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 NaBr +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>19. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Sn +\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 Al +\u00a0NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p8\" class=\"para\">20. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Mg +\u00a0HCl \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 HI +\u00a0Br<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p13\" class=\"para\">21. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0NaOH \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p14\" class=\"para\">22. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Ca(C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0HNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> +\u00a0Sr(NO<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p15\" class=\"para\">23. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 K<sub class=\"subscript\">2<\/sub>O +\u00a0MgCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p16\" class=\"para\">24. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Sn(OH)<sub class=\"subscript\">2<\/sub> +\u00a0FeBr<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 CsNO<sub class=\"subscript\">3<\/sub> +\u00a0KCl \u2192\u00a0?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p18\" class=\"para\">25. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> +\u00a0Sr(NO<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p20\" class=\"para\">26. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 KC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> +\u00a0Li<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 KOH +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p id=\"ball-ch04_s02_qs01_p19\" class=\"para\">27. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p style=\"padding-left: 30px;\">a.\u00a0 K<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a0SrCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p style=\"padding-left: 30px;\">b.\u00a0 NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<div class=\"question\">\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25099\">Show Answers to Select Questions<\/span><\/p>\n<div id=\"q25099\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>1.\u00a0<\/strong> a) \u00a0not composition, b) \u00a0composition<\/p>\n<p><strong>3.\u00a0 <\/strong>a) \u00a0composition, b) \u00a0composition<\/p>\n<div class=\"answer\">\n<p><strong>5.\u00a0 <\/strong>a) \u00a0not decomposition, b) \u00a0decomposition<\/p>\n<div class=\"answer\">\n<p><strong>7.\u00a0 <\/strong>a) \u00a0not decomposition, b) \u00a0decomposition<\/p>\n<div class=\"answer\">\n<p><strong>9.\u00a0 <\/strong>a) \u00a0combustion, b) \u00a0combustion<\/p>\n<div class=\"answer\">\n<p><strong>11. <\/strong>a) \u00a0combustion, b) \u00a0combustion<\/p>\n<div class=\"answer\">\n<p><strong>13. <\/strong>Yes; 2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O (answers will vary)<\/p>\n<p><strong>15. <\/strong>a) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a06 O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a05 H<sub class=\"subscript\">2<\/sub>O,\u00a0\u00a0 b) \u00a04 CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a03 O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O +\u00a02 N<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div>\n<p><strong>17. <\/strong>a) \u00a0Zn +\u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0Fe,\u00a0\u00a0\u00a0 b) \u00a03 F<sub class=\"subscript\">2<\/sub> +\u00a02 FeI<sub class=\"subscript\">3<\/sub> \u2192\u00a03 I<sub class=\"subscript\">2<\/sub> +\u00a02 FeF<sub class=\"subscript\">3<\/sub><\/p>\n<p><strong>19. <\/strong>a) \u00a0Sn +\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0SnSO<sub class=\"subscript\">4<\/sub> +\u00a0H<sub class=\"subscript\">2,\u00a0\u00a0\u00a0 <\/sub>b) \u00a02 Al +\u00a03 NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a02 AlBr<sub class=\"subscript\">3<\/sub> +\u00a03 Ni<\/p>\n<p><strong>21. <\/strong>a) \u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 NaOH \u2192\u00a0Zn(OH)<sub class=\"subscript\">2<\/sub> +\u00a02 NaNO<sub class=\"subscript\">3,\u00a0\u00a0\u00a0 <\/sub>b) \u00a02 HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192\u00a02 NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>S<\/p>\n<p><strong>23. <\/strong>a) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 KBr \u2192\u00a0PbBr<sub class=\"subscript\">2<\/sub> +\u00a02 KNO<sub class=\"subscript\">3<\/sub><sub class=\"subscript\">,\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sub>b) \u00a0K<sub class=\"subscript\">2<\/sub>O +\u00a0MgCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0K<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> +\u00a0MgO<\/p>\n<p><strong>27. <\/strong>a) \u00a02 K<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a03 SrCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0Sr<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>(s) +\u00a06 KCl,\u00a0\u00a0\u00a0 b) \u00a02 NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NaCl +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>(s)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p class=\"para\">\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-120\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":23485,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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