{"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":"2017-12-14T21:26:29","modified_gmt":"2017-12-14T21:26:29","slug":"types-of-chemical-reactions-single-and-double-displacement-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/types-of-chemical-reactions-single-and-double-displacement-reactions\/","title":{"raw":"Types of Chemical Reactions: Single- and Double-Displacement Reactions","rendered":"Types of Chemical Reactions: Single- and Double-Displacement Reactions"},"content":{"raw":"<div class=\"section\" id=\"ball-ch04_s02\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch04_s02_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch04_s02_l01\"><li>Recognize chemical reactions as single-replacement reactions and double-replacement reactions.<\/li>\n\t<li>Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.<\/li>\n<\/ol><\/div>\n<\/div>\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 certain types of chemical reactions that allow us to predict what the products of the reaction will be.<\/p>\n<p id=\"ball-ch04_s02_p02\" class=\"para editable block\">A <span class=\"margin_term\"><a class=\"glossterm\">single-replacement reaction<\/a><\/span>\u00a0is 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<span class=\"informalequation block\"><span class=\"mathphrase\">2 HCl(aq) +\u00a0Zn(s) \u2192\u00a0ZnCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span>\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<span class=\"informalequation block\"><span class=\"mathphrase\">2 NaCl(aq) +\u00a0F<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a02 NaF(s) +\u00a0Cl<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span>\n<p id=\"ball-ch04_s02_p04\" class=\"para editable block\">Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.<\/p>\n<p id=\"ball-ch04_s02_p05\" class=\"para editable block\">Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the <em class=\"emphasis\">halogens<\/em> and are in the next-to-last column on the periodic table (see <a class=\"xref\" href=\"#ball-ch04_s02_f01\">Figure 4.1 \"Halogens on the Periodic Table\"<\/a>). The elements on top of the column will replace the elements below them on the periodic table but not the other way around. Thus, the reaction represented by<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">CaI<sub class=\"subscript\">2<\/sub>(s) +\u00a0Cl<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a0CaCl<sub class=\"subscript\">2<\/sub>(s) +\u00a0I<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span>\n<p id=\"ball-ch04_s02_p06\" class=\"para editable block\">will occur, but the reaction<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">CaF<sub class=\"subscript\">2<\/sub>(s) +\u00a0Br<sub class=\"subscript\">2<\/sub>(\u2113) \u2192\u00a0CaBr<sub class=\"subscript\">2<\/sub>(s) +\u00a0F<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span>\n<p id=\"ball-ch04_s02_p07\" class=\"para editable block\">will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry.<\/p>\n\n<div class=\"figure large editable block\" id=\"ball-ch04_s02_f01\">\n<p class=\"title\"><span class=\"title-prefix\">Figure 4.1<\/span> Halogens on the Periodic Table<\/p>\n<p class=\"title\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Halogens.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212619\/Halogens-1.png\" alt=\"Halogens\" width=\"600\" height=\"357\" class=\"alignnone size-full wp-image-4637\"\/><\/a><\/p>\n\u00a0\n<p class=\"para\">The halogens are the elements in the next-to-last column on the periodic table.<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 2<\/h3>\n<p id=\"ball-ch04_s02_p08\" class=\"para\">Will a single-replacement reaction occur? If so, identify the products.<\/p>\n\n<ol id=\"ball-ch04_s02_l02\" class=\"orderedlist\"><li>MgCl<sub class=\"subscript\">2<\/sub> +\u00a0I<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/li>\n\t<li>CaBr<sub class=\"subscript\">2<\/sub> +\u00a0F<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/li>\n<\/ol><p class=\"simpara\">Solution<\/p>\n\n<ol id=\"ball-ch04_s02_l03\" class=\"orderedlist\"><li>Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.<\/li>\n\t<li>Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF<sub class=\"subscript\">2<\/sub> and Br<sub class=\"subscript\">2<\/sub>.<\/li>\n<\/ol><p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p09\" class=\"para\">Will a single-replacement reaction occur? If so, identify the products.<\/p>\n<p id=\"ball-ch04_s02_p10\" class=\"para\">FeI<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p11\" class=\"para\">Yes; FeCl<sub class=\"subscript\">2<\/sub> and I<sub class=\"subscript\">2<\/sub><\/p>\n\n<\/div>\n<p id=\"ball-ch04_s02_p12\" class=\"para editable block\">Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds\u2014simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the <span class=\"margin_term\"><a class=\"glossterm\">activity series<\/a><\/span>\u00a0does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.<\/p>\n\n<div class=\"callout editable block\" id=\"ball-ch04_s02_n03\">\n<h3 class=\"title\">Activity Series for Cation Replacement in Single-Replacement Reactions<\/h3>\n<ul id=\"ball-ch04_s02_l04\" class=\"itemizedlist\"><li>Li<\/li>\n\t<li>K<\/li>\n\t<li>Ba<\/li>\n\t<li>Sr<\/li>\n\t<li>Ca<\/li>\n\t<li>Na<\/li>\n\t<li>Mg<\/li>\n\t<li>Al<\/li>\n\t<li>Mn<\/li>\n\t<li>Zn<\/li>\n\t<li>Cr<\/li>\n\t<li>Fe<\/li>\n\t<li>Ni<\/li>\n\t<li>Sn<\/li>\n\t<li>Pb<\/li>\n\t<li>H<sub class=\"subscript\">2<\/sub><\/li>\n\t<li>Cu<\/li>\n\t<li>Hg<\/li>\n\t<li>Ag<\/li>\n\t<li>Pd<\/li>\n\t<li>Pt<\/li>\n\t<li>Au<\/li>\n<\/ul><\/div>\n<p id=\"ball-ch04_s02_p13\" class=\"para editable block\">Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.<\/p>\n\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch04_s02_p14\" class=\"para\">Use the activity series to predict the products, if any, of each equation.<\/p>\n\n<ol id=\"ball-ch04_s02_l05\" class=\"orderedlist\"><li>FeCl<sub class=\"subscript\">2<\/sub> +\u00a0Zn \u2192\u00a0?<\/li>\n\t<li>HNO<sub class=\"subscript\">3<\/sub> +\u00a0Au \u2192\u00a0?<\/li>\n<\/ol><p class=\"simpara\">Solution<\/p>\n\n<ol id=\"ball-ch04_s02_l06\" class=\"orderedlist\"><li>Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl<sub class=\"subscript\">2<\/sub> and Fe.<\/li>\n\t<li>Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.<\/li>\n<\/ol><p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p15\" class=\"para\">Use the activity series to predict the products, if any, of this equation.<\/p>\n<p id=\"ball-ch04_s02_p16\" class=\"para\">AlPO<sub class=\"subscript\">4<\/sub> +\u00a0Mg \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p17\" class=\"para\">Mg<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> and Al<\/p>\n\n<\/div>\n<p id=\"ball-ch04_s02_p18\" class=\"para editable block\">A <span class=\"margin_term\"><a class=\"glossterm\">double-replacement reaction<\/a><\/span>\u00a0occurs 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<span class=\"informalequation block\"><span class=\"mathphrase\">CuCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a02 AgNO<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0Cu(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a02 AgCl(s)<\/span><\/span>\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\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"ball-ch04_s02_p20\" class=\"para\">Predict the products of this double-replacement equation: BaCl<sub class=\"subscript\">2<\/sub> +\u00a0Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch04_s02_p21\" class=\"para\">Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO<sub class=\"subscript\">4<\/sub> and NaCl.<\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p22\" class=\"para\">Predict the products of this double-replacement equation: KBr +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p23\" class=\"para\">KNO<sub class=\"subscript\">3<\/sub> and AgBr<\/p>\n\n<\/div>\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 <span class=\"margin_term\"><a class=\"glossterm\">precipitation reaction<\/a><\/span>\u00a0occurs 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 <span class=\"margin_term\"><a class=\"glossterm\">precipitate<\/a><\/span>. 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 <span class=\"margin_term\"><a class=\"glossterm\">solubility rules<\/a><\/span>, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). <a class=\"xref\" href=\"#ball-ch04_s02_t01\">Table 4.1 \"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 4.1 \"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>\n\n<div class=\"table block\" id=\"ball-ch04_s02_t01\">\n<p class=\"title\"><span class=\"title-prefix\">Table 4.1<\/span> Some Useful Solubility Rules<\/p>\n\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\"><tbody><tr><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><tr><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><tr><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><tr><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><tr><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><tr><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><tr><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><tr><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><\/tbody><\/table><\/div>\n<p id=\"ball-ch04_s02_p26\" class=\"para editable block\">For example, consider the possible double-replacement reaction between Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub>. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag<sup class=\"superscript\">+<\/sup>, Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, and Pb<sup class=\"superscript\">2+<\/sup>, which are not being considered here. Therefore, Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub> are both soluble. The possible double-replacement reaction products are NaCl and SrSO<sub class=\"subscript\">4<\/sub>. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO<sub class=\"subscript\">4<\/sub>? Compounds of the sulfate ion are generally soluble, but Sr<sup class=\"superscript\">2+<\/sup> is an exception: we expect it to be insoluble\u2014a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) \u2192\u00a02 NaCl(aq) +\u00a0SrSO<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span>\n<p id=\"ball-ch04_s02_p27\" class=\"para editable block\">You would expect to see a visual change corresponding to SrSO<sub class=\"subscript\">4<\/sub> precipitating out of solution (<a class=\"xref\" href=\"#ball-ch04_s02_f02\">Figure 4.2 \"Double-Replacement Reactions\"<\/a>).<\/p>\n\n<div class=\"figure large medium-height editable block\" id=\"ball-ch04_s02_f02\">\n<p class=\"title\"><span class=\"title-prefix\">Figure 4.2<\/span> Double-Replacement Reactions<\/p>\n<p class=\"title\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Double-Replacement-Reaction.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212627\/Double-Replacement-Reaction-1.png\" alt=\"Double Replacement Reaction\" width=\"400\" height=\"426\" class=\"alignnone wp-image-4639\"\/><\/a><\/p>\n\u00a0\n<p class=\"para\">Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution.<\/p>\n\n<div class=\"copyright\">\n<p class=\"para\">Source: Photo courtesy of Choij, <a class=\"link\" href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Copper_solution.jpg\" target=\"_blank\">http:\/\/commons.wikimedia.org\/wiki\/File:Copper_solution.jpg<\/a>.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 5<\/h3>\n<p id=\"ball-ch04_s02_p28\" class=\"para\">Will a double-replacement reaction occur? If so, identify the products.<\/p>\n\n<ol id=\"ball-ch04_s02_l07\" class=\"orderedlist\"><li>Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?<\/li>\n\t<li>NaOH +\u00a0FeCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/li>\n<\/ol><p class=\"simpara\">Solution<\/p>\n\n<ol id=\"ball-ch04_s02_l08\" class=\"orderedlist\"><li>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.<\/li>\n\t<li>\n<p class=\"para\">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<span class=\"informalequation\"><span class=\"mathphrase\">2NaOH(aq) +\u00a0FeCl<sub class=\"subscript\">2<\/sub>(aq) \u2192\u00a02NaCl(aq) +\u00a0Fe(OH)<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/li>\n<\/ol><p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p29\" class=\"para\">Will a double-replacement equation occur? If so, identify the products.<\/p>\n<p id=\"ball-ch04_s02_p30\" class=\"para\">Sr(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KCl \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p31\" class=\"para\">No reaction; all possible products are soluble.<\/p>\n\n<\/div>\n<div class=\"key_takeaways editable block\" id=\"ball-ch04_s02_n07\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch04_s02_l09\" class=\"itemizedlist\"><li>A single-replacement reaction replaces one element for another in a compound.<\/li>\n\t<li>The periodic table or an activity series can help predict whether single-replacement reactions occur.<\/li>\n\t<li>A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.<\/li>\n\t<li>A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.<\/li>\n\t<li>Solubility rules are used to predict whether some double-replacement reactions will occur.<\/li>\n<\/ul><\/div>\n\u00a0\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch04_s02_qs01\">\n<h3 class=\"title\">\n\n\nWhat are the general characteristics that help you recognize single-replacement reactions?\n\n<\/h3><\/div>\n\t\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p3\" class=\"para\">What are the general characteristics that help you recognize double-replacement reactions?<\/p>\n\n<\/div>\n\t\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p5\" class=\"para\">Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p class=\"para\">a) \u00a0Zn +\u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n\n<\/div>\n\nb) \u00a0F<sub class=\"subscript\">2<\/sub> +\u00a0FeI<sub class=\"subscript\">3<\/sub> \u2192\u00a0?\n\n4. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\n\na) Li +\u00a0MgSO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?\n\nb) \u00a0NaBr +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\n5. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.\n\na) \u00a0Sn +\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?\n\nb) \u00a0Al +\u00a0NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p8\" class=\"para\">6. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\na) \u00a0Mg +\u00a0HCl \u2192\u00a0?\n\nb) \u00a0HI +\u00a0Br<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\n7. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.\n\n<\/div>\na) \u00a0FeCl<sub class=\"subscript\">2<\/sub> +\u00a0Br<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\nb) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> +\u00a0Al \u2192\u00a0?\n\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p10\" class=\"para\">8. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0Zn +\u00a0Fe<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\nb) \u00a0Ag +\u00a0HNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p11\" class=\"para\">9. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0NaI +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\nb) \u00a0AgCl +\u00a0Au \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p12\" class=\"para\">10. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0Pt +\u00a0H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?\n\nb) \u00a0Li +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0? (Hint: treat H<sub class=\"subscript\">2<\/sub>O as if it were composed of H<sup class=\"superscript\">+<\/sup> and OH<sup class=\"superscript\">\u2212<\/sup> ions.)\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p13\" class=\"para\">11. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\na) \u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0NaOH \u2192\u00a0?\n\nb) \u00a0HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p14\" class=\"para\">12. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\na) \u00a0Ca(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?\n\nb) \u00a0Na<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?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p15\" class=\"para\">13. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\na) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?\n\nb) \u00a0K<sub class=\"subscript\">2<\/sub>O +\u00a0MgCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p16\" class=\"para\">14. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\na) \u00a0Sn(OH)<sub class=\"subscript\">2<\/sub> +\u00a0FeBr<sub class=\"subscript\">3<\/sub> \u2192\u00a0?\n\nb) \u00a0CsNO<sub class=\"subscript\">3<\/sub> +\u00a0KCl \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p17\" class=\"para\">15. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?\n\nb) \u00a0K<sub class=\"subscript\">2<\/sub>O +\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p18\" class=\"para\">16. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0Na<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?\n\nb) \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?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p19\" class=\"para\">17. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0K<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a0SrCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\nb) \u00a0NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?\n\n<\/div>\n\u00a0\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p20\" class=\"para\">18. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\na) \u00a0KC<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?\n\nb) \u00a0KOH +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?\n\n<\/div>\n<\/div>\n<strong>Answers<\/strong>\n<div class=\"qandaset block\" id=\"ball-ch04_s02_qs01_ans\">\n\n<strong>1.\u00a0<\/strong>One element replaces another element in a compound.\n\n<strong>3.<\/strong>\n\na) \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\n\nb) \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><strong>5.<\/strong>\n\na) \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<\/sub>\n\nb) \u00a02 Al +\u00a03 NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a02 AlBr<sub class=\"subscript\">3<\/sub> +\u00a03 Ni\n\n<strong>7.<\/strong>\n\na) \u00a0No reaction occurs.\n\nb) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> +\u00a0Al \u2192\u00a0Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> +\u00a0Fe\n\n<strong>9.<\/strong>\n\na) \u00a02 NaI +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NaCl +\u00a0I<sub class=\"subscript\">2<\/sub>\n\nb) \u00a0No reaction occurs.\n\n<strong>11.<\/strong>\n\na) \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<\/sub>\n\nb) \u00a02 HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192\u00a02 NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>S\n\n<strong>13.<\/strong>\n\na) \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>\n\nb) \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\n\n<strong>15.<\/strong>\n\na) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 KBr \u2192\u00a0PbBr<sub class=\"subscript\">2<\/sub>(s) +\u00a02 KNO<sub class=\"subscript\">3<\/sub>\n\nb) \u00a0No reaction occurs.\n\n<strong>17.<\/strong>\n\na) \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\n\nb) \u00a02 NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NaCl +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>(s)\n\n<\/div>\n<\/div>","rendered":"<div class=\"section\" id=\"ball-ch04_s02\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch04_s02_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch04_s02_l01\">\n<li>Recognize chemical reactions as single-replacement reactions and double-replacement reactions.<\/li>\n<li>Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.<\/li>\n<\/ol>\n<\/div>\n<\/div>\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 certain types of chemical reactions that allow us to predict what the products of the reaction will be.<\/p>\n<p id=\"ball-ch04_s02_p02\" class=\"para editable block\">A <span class=\"margin_term\"><a class=\"glossterm\">single-replacement reaction<\/a><\/span>\u00a0is 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><span class=\"informalequation block\"><span class=\"mathphrase\">2 HCl(aq) +\u00a0Zn(s) \u2192\u00a0ZnCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/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><span class=\"informalequation block\"><span class=\"mathphrase\">2 NaCl(aq) +\u00a0F<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a02 NaF(s) +\u00a0Cl<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s02_p04\" class=\"para editable block\">Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.<\/p>\n<p id=\"ball-ch04_s02_p05\" class=\"para editable block\">Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the <em class=\"emphasis\">halogens<\/em> and are in the next-to-last column on the periodic table (see <a class=\"xref\" href=\"#ball-ch04_s02_f01\">Figure 4.1 &#8220;Halogens on the Periodic Table&#8221;<\/a>). The elements on top of the column will replace the elements below them on the periodic table but not the other way around. Thus, the reaction represented by<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">CaI<sub class=\"subscript\">2<\/sub>(s) +\u00a0Cl<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a0CaCl<sub class=\"subscript\">2<\/sub>(s) +\u00a0I<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s02_p06\" class=\"para editable block\">will occur, but the reaction<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">CaF<sub class=\"subscript\">2<\/sub>(s) +\u00a0Br<sub class=\"subscript\">2<\/sub>(\u2113) \u2192\u00a0CaBr<sub class=\"subscript\">2<\/sub>(s) +\u00a0F<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s02_p07\" class=\"para editable block\">will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry.<\/p>\n<div class=\"figure large editable block\" id=\"ball-ch04_s02_f01\">\n<p class=\"title\"><span class=\"title-prefix\">Figure 4.1<\/span> Halogens on the Periodic Table<\/p>\n<p class=\"title\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Halogens.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212619\/Halogens-1.png\" alt=\"Halogens\" width=\"600\" height=\"357\" class=\"alignnone size-full wp-image-4637\" \/><\/a><\/p>\n<p>\u00a0<\/p>\n<p class=\"para\">The halogens are the elements in the next-to-last column on the periodic table.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 2<\/h3>\n<p id=\"ball-ch04_s02_p08\" class=\"para\">Will a single-replacement reaction occur? If so, identify the products.<\/p>\n<ol id=\"ball-ch04_s02_l02\" class=\"orderedlist\">\n<li>MgCl<sub class=\"subscript\">2<\/sub> +\u00a0I<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/li>\n<li>CaBr<sub class=\"subscript\">2<\/sub> +\u00a0F<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch04_s02_l03\" class=\"orderedlist\">\n<li>Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.<\/li>\n<li>Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF<sub class=\"subscript\">2<\/sub> and Br<sub class=\"subscript\">2<\/sub>.<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p09\" class=\"para\">Will a single-replacement reaction occur? If so, identify the products.<\/p>\n<p id=\"ball-ch04_s02_p10\" class=\"para\">FeI<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p11\" class=\"para\">Yes; FeCl<sub class=\"subscript\">2<\/sub> and I<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<p id=\"ball-ch04_s02_p12\" class=\"para editable block\">Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds\u2014simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the <span class=\"margin_term\"><a class=\"glossterm\">activity series<\/a><\/span>\u00a0does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.<\/p>\n<div class=\"callout editable block\" id=\"ball-ch04_s02_n03\">\n<h3 class=\"title\">Activity Series for Cation Replacement in Single-Replacement Reactions<\/h3>\n<ul id=\"ball-ch04_s02_l04\" class=\"itemizedlist\">\n<li>Li<\/li>\n<li>K<\/li>\n<li>Ba<\/li>\n<li>Sr<\/li>\n<li>Ca<\/li>\n<li>Na<\/li>\n<li>Mg<\/li>\n<li>Al<\/li>\n<li>Mn<\/li>\n<li>Zn<\/li>\n<li>Cr<\/li>\n<li>Fe<\/li>\n<li>Ni<\/li>\n<li>Sn<\/li>\n<li>Pb<\/li>\n<li>H<sub class=\"subscript\">2<\/sub><\/li>\n<li>Cu<\/li>\n<li>Hg<\/li>\n<li>Ag<\/li>\n<li>Pd<\/li>\n<li>Pt<\/li>\n<li>Au<\/li>\n<\/ul>\n<\/div>\n<p id=\"ball-ch04_s02_p13\" class=\"para editable block\">Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch04_s02_p14\" class=\"para\">Use the activity series to predict the products, if any, of each equation.<\/p>\n<ol id=\"ball-ch04_s02_l05\" class=\"orderedlist\">\n<li>FeCl<sub class=\"subscript\">2<\/sub> +\u00a0Zn \u2192\u00a0?<\/li>\n<li>HNO<sub class=\"subscript\">3<\/sub> +\u00a0Au \u2192\u00a0?<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch04_s02_l06\" class=\"orderedlist\">\n<li>Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl<sub class=\"subscript\">2<\/sub> and Fe.<\/li>\n<li>Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p15\" class=\"para\">Use the activity series to predict the products, if any, of this equation.<\/p>\n<p id=\"ball-ch04_s02_p16\" class=\"para\">AlPO<sub class=\"subscript\">4<\/sub> +\u00a0Mg \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p17\" class=\"para\">Mg<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> and Al<\/p>\n<\/div>\n<p id=\"ball-ch04_s02_p18\" class=\"para editable block\">A <span class=\"margin_term\"><a class=\"glossterm\">double-replacement reaction<\/a><\/span>\u00a0occurs 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><span class=\"informalequation block\"><span class=\"mathphrase\">CuCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a02 AgNO<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0Cu(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a02 AgCl(s)<\/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 class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"ball-ch04_s02_p20\" class=\"para\">Predict the products of this double-replacement equation: BaCl<sub class=\"subscript\">2<\/sub> +\u00a0Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch04_s02_p21\" class=\"para\">Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO<sub class=\"subscript\">4<\/sub> and NaCl.<\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p22\" class=\"para\">Predict the products of this double-replacement equation: KBr +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p23\" class=\"para\">KNO<sub class=\"subscript\">3<\/sub> and AgBr<\/p>\n<\/div>\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 <span class=\"margin_term\"><a class=\"glossterm\">precipitation reaction<\/a><\/span>\u00a0occurs 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 <span class=\"margin_term\"><a class=\"glossterm\">precipitate<\/a><\/span>. 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 <span class=\"margin_term\"><a class=\"glossterm\">solubility rules<\/a><\/span>, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). <a class=\"xref\" href=\"#ball-ch04_s02_t01\">Table 4.1 &#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 4.1 &#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 class=\"table block\" id=\"ball-ch04_s02_t01\">\n<p class=\"title\"><span class=\"title-prefix\">Table 4.1<\/span> Some Useful Solubility Rules<\/p>\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\">\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<p id=\"ball-ch04_s02_p26\" class=\"para editable block\">For example, consider the possible double-replacement reaction between Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub>. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag<sup class=\"superscript\">+<\/sup>, Hg<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2+<\/sup>, and Pb<sup class=\"superscript\">2+<\/sup>, which are not being considered here. Therefore, Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and SrCl<sub class=\"subscript\">2<\/sub> are both soluble. The possible double-replacement reaction products are NaCl and SrSO<sub class=\"subscript\">4<\/sub>. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO<sub class=\"subscript\">4<\/sub>? Compounds of the sulfate ion are generally soluble, but Sr<sup class=\"superscript\">2+<\/sup> is an exception: we expect it to be insoluble\u2014a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) \u2192\u00a02 NaCl(aq) +\u00a0SrSO<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s02_p27\" class=\"para editable block\">You would expect to see a visual change corresponding to SrSO<sub class=\"subscript\">4<\/sub> precipitating out of solution (<a class=\"xref\" href=\"#ball-ch04_s02_f02\">Figure 4.2 &#8220;Double-Replacement Reactions&#8221;<\/a>).<\/p>\n<div class=\"figure large medium-height editable block\" id=\"ball-ch04_s02_f02\">\n<p class=\"title\"><span class=\"title-prefix\">Figure 4.2<\/span> Double-Replacement Reactions<\/p>\n<p class=\"title\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Double-Replacement-Reaction.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212627\/Double-Replacement-Reaction-1.png\" alt=\"Double Replacement Reaction\" width=\"400\" height=\"426\" class=\"alignnone wp-image-4639\" \/><\/a><\/p>\n<p>\u00a0<\/p>\n<p class=\"para\">Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution.<\/p>\n<div class=\"copyright\">\n<p class=\"para\">Source: Photo courtesy of Choij, <a class=\"link\" href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Copper_solution.jpg\" target=\"_blank\">http:\/\/commons.wikimedia.org\/wiki\/File:Copper_solution.jpg<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 5<\/h3>\n<p id=\"ball-ch04_s02_p28\" class=\"para\">Will a double-replacement reaction occur? If so, identify the products.<\/p>\n<ol id=\"ball-ch04_s02_l07\" class=\"orderedlist\">\n<li>Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?<\/li>\n<li>NaOH +\u00a0FeCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch04_s02_l08\" class=\"orderedlist\">\n<li>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.<\/li>\n<li>\n<p class=\"para\">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><span class=\"informalequation\"><span class=\"mathphrase\">2NaOH(aq) +\u00a0FeCl<sub class=\"subscript\">2<\/sub>(aq) \u2192\u00a02NaCl(aq) +\u00a0Fe(OH)<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s02_p29\" class=\"para\">Will a double-replacement equation occur? If so, identify the products.<\/p>\n<p id=\"ball-ch04_s02_p30\" class=\"para\">Sr(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KCl \u2192\u00a0?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s02_p31\" class=\"para\">No reaction; all possible products are soluble.<\/p>\n<\/div>\n<div class=\"key_takeaways editable block\" id=\"ball-ch04_s02_n07\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch04_s02_l09\" class=\"itemizedlist\">\n<li>A single-replacement reaction replaces one element for another in a compound.<\/li>\n<li>The periodic table or an activity series can help predict whether single-replacement reactions occur.<\/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<p>\u00a0<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch04_s02_qs01\">\n<h3 class=\"title\">\n<p>What are the general characteristics that help you recognize single-replacement reactions?<\/p>\n<\/h3>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p3\" class=\"para\">What are the general characteristics that help you recognize double-replacement reactions?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p5\" class=\"para\">Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p class=\"para\">a) \u00a0Zn +\u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p>b) \u00a0F<sub class=\"subscript\">2<\/sub> +\u00a0FeI<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p>4. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) Li +\u00a0MgSO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0NaBr +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>5. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) \u00a0Sn +\u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0Al +\u00a0NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p8\" class=\"para\">6. \u00a0Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) \u00a0Mg +\u00a0HCl \u2192\u00a0?<\/p>\n<p>b) \u00a0HI +\u00a0Br<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>7. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<\/div>\n<p>a) \u00a0FeCl<sub class=\"subscript\">2<\/sub> +\u00a0Br<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> +\u00a0Al \u2192\u00a0?<\/p>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p10\" class=\"para\">8. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0Zn +\u00a0Fe<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0Ag +\u00a0HNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p11\" class=\"para\">9. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0NaI +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0AgCl +\u00a0Au \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p12\" class=\"para\">10. \u00a0Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0Pt +\u00a0H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0Li +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0? (Hint: treat H<sub class=\"subscript\">2<\/sub>O as if it were composed of H<sup class=\"superscript\">+<\/sup> and OH<sup class=\"superscript\">\u2212<\/sup> ions.)<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p13\" class=\"para\">11. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) \u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0NaOH \u2192\u00a0?<\/p>\n<p>b) \u00a0HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p14\" class=\"para\">12. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) \u00a0Ca(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>b) \u00a0Na<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<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p15\" class=\"para\">13. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?<\/p>\n<p>b) \u00a0K<sub class=\"subscript\">2<\/sub>O +\u00a0MgCO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p16\" class=\"para\">14. \u00a0Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.<\/p>\n<p>a) \u00a0Sn(OH)<sub class=\"subscript\">2<\/sub> +\u00a0FeBr<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0CsNO<sub class=\"subscript\">3<\/sub> +\u00a0KCl \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p17\" class=\"para\">15. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a0KBr \u2192\u00a0?<\/p>\n<p>b) \u00a0K<sub class=\"subscript\">2<\/sub>O +\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p18\" class=\"para\">16. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0Na<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>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<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p19\" class=\"para\">17. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0K<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a0SrCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<p>b) \u00a0NaOH +\u00a0MgCl<sub class=\"subscript\">2<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s02_qs01_p20\" class=\"para\">18. \u00a0Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.<\/p>\n<p>a) \u00a0KC<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>b) \u00a0KOH +\u00a0AgNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0?<\/p>\n<\/div>\n<\/div>\n<p><strong>Answers<\/strong><\/p>\n<div class=\"qandaset block\" id=\"ball-ch04_s02_qs01_ans\">\n<p><strong>1.\u00a0<\/strong>One element replaces another element in a compound.<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>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<\/p>\n<p>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><strong>5.<\/strong><\/p>\n<p>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<\/sub><\/p>\n<p>b) \u00a02 Al +\u00a03 NiBr<sub class=\"subscript\">2<\/sub> \u2192\u00a02 AlBr<sub class=\"subscript\">3<\/sub> +\u00a03 Ni<\/p>\n<p><strong>7.<\/strong><\/p>\n<p>a) \u00a0No reaction occurs.<\/p>\n<p>b) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> +\u00a0Al \u2192\u00a0Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub> +\u00a0Fe<\/p>\n<p><strong>9.<\/strong><\/p>\n<p>a) \u00a02 NaI +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NaCl +\u00a0I<sub class=\"subscript\">2<\/sub><\/p>\n<p>b) \u00a0No reaction occurs.<\/p>\n<p><strong>11.<\/strong><\/p>\n<p>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<\/sub><\/p>\n<p>b) \u00a02 HCl +\u00a0Na<sub class=\"subscript\">2<\/sub>S \u2192\u00a02 NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>S<\/p>\n<p><strong>13.<\/strong><\/p>\n<p>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><\/p>\n<p>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>15.<\/strong><\/p>\n<p>a) \u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 KBr \u2192\u00a0PbBr<sub class=\"subscript\">2<\/sub>(s) +\u00a02 KNO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0No reaction occurs.<\/p>\n<p><strong>17.<\/strong><\/p>\n<p>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<\/p>\n<p>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\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. Ball\",\"organization\":\"BCCampus\",\"url\":\"https:\/\/opentextbc.ca\/introductorychemistry\/\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Download this book for free at http:\/\/open.bccampus.ca\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-120","chapter","type-chapter","status-publish","hentry"],"part":114,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters\/120","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/users\/23485"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters\/120\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/parts\/114"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters\/120\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/media?parent=120"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=120"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/contributor?post=120"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/license?post=120"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}