{"id":117,"date":"2017-12-14T21:26:18","date_gmt":"2017-12-14T21:26:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/the-chemical-equation\/"},"modified":"2017-12-14T21:26:18","modified_gmt":"2017-12-14T21:26:18","slug":"the-chemical-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/the-chemical-equation\/","title":{"raw":"The Chemical Equation","rendered":"The Chemical Equation"},"content":{"raw":"<div>\n<div class=\"section\" id=\"ball-ch04_s01\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch04_s01_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch04_s01_l01\"><li>Define <em>chemical equation<\/em>.<\/li>\n\t<li>Identify the parts of a chemical equation.<\/li>\n<\/ol><\/div>\n<\/div>\n<p id=\"ball-ch04_s01_p01\" class=\"para editable block\">A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">hydrogen reacts with oxygen to make water<\/span><\/span>\n<p id=\"ball-ch04_s01_p02\" class=\"para editable block\">We can represent this chemical change more succinctly as<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">hydrogen +\u00a0oxygen \u2192\u00a0water<\/span><\/span>\n<p id=\"ball-ch04_s01_p03\" class=\"para editable block\">where the +\u00a0sign means that the two substances interact chemically with each other and the \u2192\u00a0symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p04\" class=\"para editable block\">This is an example of a <span class=\"margin_term\"><a class=\"glossterm\">chemical equation<\/a><\/span>, which is a concise way of representing a chemical reaction. The initial substances are called <span class=\"margin_term\"><a class=\"glossterm\">reactants<\/a><\/span>, and the final substances are called <span class=\"margin_term\"><a class=\"glossterm\">products<\/a><\/span>.<\/p>\n<p id=\"ball-ch04_s01_p05\" class=\"para editable block\">Unfortunately, it is also an <em class=\"emphasis\">incomplete<\/em> chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.<\/p>\n<p id=\"ball-ch04_s01_p06\" class=\"para editable block\">What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No; you <em class=\"emphasis\">cannot<\/em> change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance. What you <em class=\"emphasis\">can<\/em> do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side.<\/p>\n<p id=\"ball-ch04_s01_p07\" class=\"para editable block\">To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p08\" class=\"para editable block\">The 2 in front of the formula for water is called a <span class=\"margin_term\"><a class=\"glossterm\">coefficient<\/a><\/span>. Now there is the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem\u2014simply go back to the reactant side of the equation and add a coefficient in front of the H<sub class=\"subscript\">2<\/sub>. The coefficient that works is 2:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p09\" class=\"para editable block\">There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is <span class=\"margin_term\"><a class=\"glossterm\">balanced<\/a><\/span>. All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more \u201cback and forth\u201d work when balancing a chemical equation.<\/p>\n\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 1<\/h3>\n<p id=\"ball-ch04_s01_p10\" class=\"para\">Write and balance the chemical equation for each given chemical reaction.<\/p>\n\n<ol id=\"ball-ch04_s01_l02\" class=\"orderedlist\"><li>Hydrogen and chlorine react to make HCl.<\/li>\n\t<li>Ethane, C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub>, reacts with oxygen to make carbon dioxide and water.<\/li>\n<\/ol><p class=\"simpara\">Solution<\/p>\n\n<ol id=\"ball-ch04_s01_l03\" class=\"orderedlist\"><li>\n<p class=\"para\">Let us start by simply writing a chemical equation in terms of the formulas of the substances, remembering that both elemental hydrogen and chlorine are diatomic:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0HCl<\/span><\/span>\n<p id=\"ball-ch04_s01_p11\" class=\"para\">There are two hydrogen atoms and two chlorine atoms in the reactants and one of each atom in the product. We can fix this by including the coefficient 2 on the product side:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 HCl<\/span><\/span>\n<p id=\"ball-ch04_s01_p12\" class=\"para\">Now there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced.<\/p>\n<\/li>\n\t<li>\n<p class=\"para\">Start by writing the chemical equation in terms of the substances involved:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0CO<sub class=\"subscript\">2<\/sub> +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p13\" class=\"para\">We have two carbon atoms on the left, so we need two carbon dioxide molecules on the product side, so that each side has two carbon atoms; that element is balanced. We have six hydrogen atoms in the reactants, so we need six hydrogen atoms in the products. We can get this by having three water molecules:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 CO<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p14\" class=\"para\">Now we have seven oxygen atoms in the products (four from the CO<sub class=\"subscript\">2<\/sub> and three from the H<sub class=\"subscript\">2<\/sub>O). That means we need seven oxygen atoms in the reactants. However, because oxygen is a diatomic molecule, we can only get an even number of oxygen atoms at a time. We can achieve this by multiplying the other coefficients by 2:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p15\" class=\"para\">By multiplying everything else by 2, we don\u2019t unbalance the other elements, and we now get an even number of oxygen atoms in the product\u201414. We can get 14 oxygen atoms on the reactant side by having 7 oxygen molecules:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a07 O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch04_s01_p16\" class=\"para\">As a check, recount everything to determine that each side has the same number of atoms of each element. This chemical equation is now balanced.<\/p>\n<\/li>\n<\/ol><p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s01_p17\" class=\"para\">Write and balance the chemical equation that represents nitrogen and hydrogen reacting to produce ammonia, NH<sub class=\"subscript\">3<\/sub>.<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s01_p18\" class=\"para\">N<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub><\/p>\n\n<\/div>\n<p id=\"ball-ch04_s01_p19\" class=\"para editable block\">Many chemical equations also include phase labels for the substances: (s) for solid, (\u2113) for liquid, (g) for gas, and (aq) for aqueous (i.e., dissolved in water). Special conditions, such as temperature, may also be listed above the arrow. For example,<\/p>\n<span class=\"informalequation block\">2 NaHCO<sub>3<\/sub>(s)\u00a0\u2192200\u00b0C\u00a0Na<sub>2<\/sub>CO<sub>3<\/sub>(s) + CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(\u2113)<\/span>\n\n\u00a0\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch04_s01_l04\" class=\"itemizedlist\"><li>A chemical equation is a concise description of a chemical reaction.<\/li>\n\t<li>Proper chemical equations are balanced.<\/li>\n<\/ul><\/div>\n\u00a0\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<ol id=\"ball-ch04_s01_qs01_qd01\" class=\"qandadiv\"><li id=\"ball-ch04_s01_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p1\" class=\"para\">From the statement \u201cnitrogen and hydrogen react to produce ammonia,\u201d identify the reactants and the products.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p3\" class=\"para\">From the statement \u201csodium metal reacts with water to produce sodium hydroxide and hydrogen,\u201d identify the reactants and the products.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p5\" class=\"para\">From the statement \u201cmagnesium hydroxide reacts with nitric acid to produce magnesium nitrate and water,\u201d identify the reactants and the products.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p7\" class=\"para\">From the statement \u201cpropane reacts with oxygen to produce carbon dioxide and water,\u201d identify the reactants and the products.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p9\" class=\"para\">Write and balance the chemical equation described by Exercise 1.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p11\" class=\"para\">Write and balance the chemical equation described by Exercise 2.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p13\" class=\"para\">Write and balance the chemical equation described by Exercise 3.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p15\" class=\"para\">Write and balance the chemical equation described by Exercise 4. The formula for propane is C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub>.<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p17\" class=\"para\">Balance: ___NaClO<sub class=\"subscript\">3<\/sub> \u2192\u00a0___NaCl +\u00a0___O<sub class=\"subscript\">2<\/sub><\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p19\" class=\"para\">Balance: ___N<sub class=\"subscript\">2<\/sub> +\u00a0___H<sub class=\"subscript\">2<\/sub> \u2192\u00a0___N<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub><\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa11\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p21\" class=\"para\">Balance: ___Al +\u00a0___O<sub class=\"subscript\">2<\/sub> \u2192\u00a0___Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub><\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa12\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p23\" class=\"para\">Balance: ___C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a0___O<sub class=\"subscript\">2<\/sub> \u2192\u00a0___CO<sub class=\"subscript\">2<\/sub> +\u00a0___H<sub class=\"subscript\">2<\/sub>O<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa13\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p25\" class=\"para\">How would you write the balanced chemical equation in Exercise 10 if all substances were gases?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch04_s01_qs01_qd01_qa14\" class=\"qandaentry\">How would you write the balanced chemical equation in Exercise 12 if all the substances except water were gases and water itself were a liquid?<\/li>\n<\/ol><p class=\"para\"><strong>Answers<\/strong><\/p>\n\n<div class=\"qandaset block\" id=\"ball-ch04_s01_qs01_ans\">\n\n<strong>1.<\/strong>\n\n<\/div>\n<div class=\"qandaset block\">\n\nreactants: nitrogen and hydrogen; product: ammonia\n\n<strong>3.<\/strong>\n\nreactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water\n\n<strong>5.<\/strong>\n\nN<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub><strong>7.<\/strong>\n\nMg(OH)<sub class=\"subscript\">2<\/sub> +\u00a02 HNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0Mg(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O\n\n<strong>9.<\/strong>\n\n2 NaClO<sub class=\"subscript\">3<\/sub> \u2192\u00a02 NaCl +\u00a03 O<sub class=\"subscript\">2<\/sub><strong>11.<\/strong>\n\n4Al +\u00a03O<sub class=\"subscript\">2<\/sub> \u2192\u00a02Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub><strong>13.<\/strong>\n\nN<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)\n\n<\/div>\n<\/div>\n\u00a0\n<div class=\"qandaset block\" id=\"ball-ch04_s01_qs01\">\n<div class=\"qandaset block\">\n<\/div>\n<\/div><\/div><\/div>","rendered":"<div>\n<div class=\"section\" id=\"ball-ch04_s01\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch04_s01_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch04_s01_l01\">\n<li>Define <em>chemical equation<\/em>.<\/li>\n<li>Identify the parts of a chemical equation.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"ball-ch04_s01_p01\" class=\"para editable block\">A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">hydrogen reacts with oxygen to make water<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p02\" class=\"para editable block\">We can represent this chemical change more succinctly as<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">hydrogen +\u00a0oxygen \u2192\u00a0water<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p03\" class=\"para editable block\">where the +\u00a0sign means that the two substances interact chemically with each other and the \u2192\u00a0symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p04\" class=\"para editable block\">This is an example of a <span class=\"margin_term\"><a class=\"glossterm\">chemical equation<\/a><\/span>, which is a concise way of representing a chemical reaction. The initial substances are called <span class=\"margin_term\"><a class=\"glossterm\">reactants<\/a><\/span>, and the final substances are called <span class=\"margin_term\"><a class=\"glossterm\">products<\/a><\/span>.<\/p>\n<p id=\"ball-ch04_s01_p05\" class=\"para editable block\">Unfortunately, it is also an <em class=\"emphasis\">incomplete<\/em> chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.<\/p>\n<p id=\"ball-ch04_s01_p06\" class=\"para editable block\">What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No; you <em class=\"emphasis\">cannot<\/em> change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance. What you <em class=\"emphasis\">can<\/em> do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side.<\/p>\n<p id=\"ball-ch04_s01_p07\" class=\"para editable block\">To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p08\" class=\"para editable block\">The 2 in front of the formula for water is called a <span class=\"margin_term\"><a class=\"glossterm\">coefficient<\/a><\/span>. Now there is the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem\u2014simply go back to the reactant side of the equation and add a coefficient in front of the H<sub class=\"subscript\">2<\/sub>. The coefficient that works is 2:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p09\" class=\"para editable block\">There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is <span class=\"margin_term\"><a class=\"glossterm\">balanced<\/a><\/span>. All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more \u201cback and forth\u201d work when balancing a chemical equation.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 1<\/h3>\n<p id=\"ball-ch04_s01_p10\" class=\"para\">Write and balance the chemical equation for each given chemical reaction.<\/p>\n<ol id=\"ball-ch04_s01_l02\" class=\"orderedlist\">\n<li>Hydrogen and chlorine react to make HCl.<\/li>\n<li>Ethane, C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub>, reacts with oxygen to make carbon dioxide and water.<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch04_s01_l03\" class=\"orderedlist\">\n<li>\n<p class=\"para\">Let us start by simply writing a chemical equation in terms of the formulas of the substances, remembering that both elemental hydrogen and chlorine are diatomic:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a0HCl<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p11\" class=\"para\">There are two hydrogen atoms and two chlorine atoms in the reactants and one of each atom in the product. We can fix this by including the coefficient 2 on the product side:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> \u2192\u00a02 HCl<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p12\" class=\"para\">Now there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced.<\/p>\n<\/li>\n<li>\n<p class=\"para\">Start by writing the chemical equation in terms of the substances involved:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a0CO<sub class=\"subscript\">2<\/sub> +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p13\" class=\"para\">We have two carbon atoms on the left, so we need two carbon dioxide molecules on the product side, so that each side has two carbon atoms; that element is balanced. We have six hydrogen atoms in the reactants, so we need six hydrogen atoms in the products. We can get this by having three water molecules:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a02 CO<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p14\" class=\"para\">Now we have seven oxygen atoms in the products (four from the CO<sub class=\"subscript\">2<\/sub> and three from the H<sub class=\"subscript\">2<\/sub>O). That means we need seven oxygen atoms in the reactants. However, because oxygen is a diatomic molecule, we can only get an even number of oxygen atoms at a time. We can achieve this by multiplying the other coefficients by 2:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p15\" class=\"para\">By multiplying everything else by 2, we don\u2019t unbalance the other elements, and we now get an even number of oxygen atoms in the product\u201414. We can get 14 oxygen atoms on the reactant side by having 7 oxygen molecules:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a07 O<sub class=\"subscript\">2<\/sub> \u2192\u00a04 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s01_p16\" class=\"para\">As a check, recount everything to determine that each side has the same number of atoms of each element. This chemical equation is now balanced.<\/p>\n<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch04_s01_p17\" class=\"para\">Write and balance the chemical equation that represents nitrogen and hydrogen reacting to produce ammonia, NH<sub class=\"subscript\">3<\/sub>.<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch04_s01_p18\" class=\"para\">N<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<p id=\"ball-ch04_s01_p19\" class=\"para editable block\">Many chemical equations also include phase labels for the substances: (s) for solid, (\u2113) for liquid, (g) for gas, and (aq) for aqueous (i.e., dissolved in water). Special conditions, such as temperature, may also be listed above the arrow. For example,<\/p>\n<p><span class=\"informalequation block\">2 NaHCO<sub>3<\/sub>(s)\u00a0\u2192200\u00b0C\u00a0Na<sub>2<\/sub>CO<sub>3<\/sub>(s) + CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(\u2113)<\/span><\/p>\n<p>\u00a0<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch04_s01_l04\" class=\"itemizedlist\">\n<li>A chemical equation is a concise description of a chemical reaction.<\/li>\n<li>Proper chemical equations are balanced.<\/li>\n<\/ul>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<ol id=\"ball-ch04_s01_qs01_qd01\" class=\"qandadiv\">\n<li id=\"ball-ch04_s01_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p1\" class=\"para\">From the statement \u201cnitrogen and hydrogen react to produce ammonia,\u201d identify the reactants and the products.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p3\" class=\"para\">From the statement \u201csodium metal reacts with water to produce sodium hydroxide and hydrogen,\u201d identify the reactants and the products.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p5\" class=\"para\">From the statement \u201cmagnesium hydroxide reacts with nitric acid to produce magnesium nitrate and water,\u201d identify the reactants and the products.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p7\" class=\"para\">From the statement \u201cpropane reacts with oxygen to produce carbon dioxide and water,\u201d identify the reactants and the products.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p9\" class=\"para\">Write and balance the chemical equation described by Exercise 1.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p11\" class=\"para\">Write and balance the chemical equation described by Exercise 2.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p13\" class=\"para\">Write and balance the chemical equation described by Exercise 3.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p15\" class=\"para\">Write and balance the chemical equation described by Exercise 4. The formula for propane is C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub>.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p17\" class=\"para\">Balance: ___NaClO<sub class=\"subscript\">3<\/sub> \u2192\u00a0___NaCl +\u00a0___O<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p19\" class=\"para\">Balance: ___N<sub class=\"subscript\">2<\/sub> +\u00a0___H<sub class=\"subscript\">2<\/sub> \u2192\u00a0___N<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa11\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p21\" class=\"para\">Balance: ___Al +\u00a0___O<sub class=\"subscript\">2<\/sub> \u2192\u00a0___Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa12\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p23\" class=\"para\">Balance: ___C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a0___O<sub class=\"subscript\">2<\/sub> \u2192\u00a0___CO<sub class=\"subscript\">2<\/sub> +\u00a0___H<sub class=\"subscript\">2<\/sub>O<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa13\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch04_s01_qs01_p25\" class=\"para\">How would you write the balanced chemical equation in Exercise 10 if all substances were gases?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch04_s01_qs01_qd01_qa14\" class=\"qandaentry\">How would you write the balanced chemical equation in Exercise 12 if all the substances except water were gases and water itself were a liquid?<\/li>\n<\/ol>\n<p class=\"para\"><strong>Answers<\/strong><\/p>\n<div class=\"qandaset block\" id=\"ball-ch04_s01_qs01_ans\">\n<p><strong>1.<\/strong><\/p>\n<\/div>\n<div class=\"qandaset block\">\n<p>reactants: nitrogen and hydrogen; product: ammonia<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water<\/p>\n<p><strong>5.<\/strong><\/p>\n<p>N<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub> \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub><strong>7.<\/strong><\/p>\n<p>Mg(OH)<sub class=\"subscript\">2<\/sub> +\u00a02 HNO<sub class=\"subscript\">3<\/sub> \u2192\u00a0Mg(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p><strong>9.<\/strong><\/p>\n<p>2 NaClO<sub class=\"subscript\">3<\/sub> \u2192\u00a02 NaCl +\u00a03 O<sub class=\"subscript\">2<\/sub><strong>11.<\/strong><\/p>\n<p>4Al +\u00a03O<sub class=\"subscript\">2<\/sub> \u2192\u00a02Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub><strong>13.<\/strong><\/p>\n<p>N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/p>\n<\/div>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"qandaset block\" id=\"ball-ch04_s01_qs01\">\n<div class=\"qandaset block\">\n<\/div>\n<\/div>\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-117\">\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":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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