{"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":"2019-01-04T04:13:44","modified_gmt":"2019-01-04T04:13:44","slug":"the-chemical-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/the-chemical-equation\/","title":{"raw":"7.2 The Chemical Equation: Balancing Chemical Equations","rendered":"7.2 The Chemical Equation: Balancing Chemical Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Derive chemical equations from narrative descriptions of chemical reactions.<\/li>\r\n \t<li>Write and balance chemical equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a <strong>chemical equation<\/strong>. Consider as an example the reaction between one methane molecule (CH<sub>4<\/sub>) and two diatomic oxygen molecules (O<sub>2<\/sub>) to produce one carbon dioxide molecule (CO<sub>2<\/sub>) and two water molecules (H<sub>2<\/sub>O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"881\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211218\/CNX_Chem_04_01_rxn21.jpg\" alt=\"This figure shows a balanced chemical equation followed below by a representation of the equation using space-filling models. The equation reads C H subscript 4 plus 2 O subscript 2 arrow C O subscript 2 plus 2 H subscript 2 O. Under the C H subscript 4, the molecule is shown with a central black sphere, representing a C atom, to which 4 smaller white spheres, representing H atoms, are distributed evenly around. All four H atoms are bonded to the central black C atom. This is followed by a plus sign. Under the 2 O subscript 2, two molecules are shown. The molecules are each composed of two red spheres bonded together. The red spheres represent O atoms. To the right of an arrow and under the C O subscript 2, appears a single molecule with a black central sphere with two red spheres bonded to the left and right. Following a plus sign and under the 2 H subscript 2 O, are two molecules, each with a central red sphere and two smaller white spheres attached to the lower right and lower left sides of the central red sphere. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms.\" width=\"881\" height=\"401\" \/> Figure 1. The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).[\/caption]\r\n\r\nThis example illustrates the fundamental aspects of any chemical equation:\r\n<ol>\r\n \t<li>The substances undergoing reaction are called <strong>reactants<\/strong>, and their formulas are placed on the left side of the equation.<\/li>\r\n \t<li>The substances generated by the reaction are called <strong>products<\/strong>, and their formulas are placed on the right sight of the equation.<\/li>\r\n \t<li>Plus signs (+) separate individual reactant and product formulas, and an arrow ([latex]\\rightarrow[\/latex]) separates the reactant and product (left and right) sides of the equation.<\/li>\r\n \t<li>The relative numbers of reactant and product species are represented by <strong>coefficients<\/strong> (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.<\/li>\r\n<\/ol>\r\nIt is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the <em>relative<\/em> numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 2). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:\r\n<ul id=\"fs-idp42981680\">\r\n \t<li><em>One<\/em> methane molecule and <em>two<\/em> oxygen molecules react to yield <em>one<\/em> carbon dioxide molecule and <em>two<\/em> water molecules.<\/li>\r\n \t<li><em>One dozen<\/em> methane molecules and <em>two dozen<\/em> oxygen molecules react to yield <em>one dozen<\/em> carbon dioxide molecules and <em>two dozen<\/em> water molecules.<\/li>\r\n \t<li><em>One mole<\/em> of methane molecules and <em>2 moles<\/em> of oxygen molecules react to yield <em>1 mole<\/em> of carbon dioxide molecules and <em>2 moles<\/em> of water molecules.<\/li>\r\n<\/ul>\r\n[caption id=\"\" align=\"alignnone\" width=\"880\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211219\/CNX_Chem_04_01_rxn31.jpg\" alt=\"This image has a left side, labeled, \u201cMixture before reaction\u201d separated by a vertical dashed line from right side labeled, \u201cMixture after reaction.\u201d On the left side of the figure, two types of molecules are illustrated with space-filling models. Six of the molecules have only two red spheres bonded together. Three of the molecules have four small white spheres evenly distributed about and bonded to a central, larger black sphere. On the right side of the dashed vertical line, two types of molecules which are different from those on the left side are shown. Six of the molecules have a central red sphere to which smaller white spheres are bonded. The white spheres are not opposite each other on the red atoms, giving the molecule a bent shape or appearance. The second molecule type has a central black sphere to which two red spheres are attached on opposite sides, resulting in a linear shape or appearance. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms. On each side of the dashed line, twelve red, three black, and twelve white spheres are present.\" width=\"880\" height=\"341\" \/> Figure 2. Regardless of the absolute number of molecules involved, the ratios between numbers of molecules are the same as that given in the chemical equation.[\/caption]\r\n<h2>Balancing Equations<\/h2>\r\nA <strong>balanced chemical is equation <\/strong>has equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element\u2019s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO<sub>2<\/sub> and H<sub>2<\/sub>O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is\r\n<p style=\"text-align: center\">[latex]\\large\\left(1{\\text{ CO}}_{2}\\text{ molecule }\\times \\frac{\\text{2 O atoms}}{{\\text{1 CO}}_{2}\\text{ molecule }}\\right)+\\left(2{\\text{ H}}_{2}\\text{O molecule }\\times \\frac{\\text{1 O atom}}{{\\text{1 H}}_{2}\\text{O molecule }}\\right)=\\text{4 O atoms}[\/latex]<\/p>\r\nThe equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{CH}}_{4}+2{\\text{ O}}_{2}\\rightarrow{\\text{CO}}_{2}+2{\\text{ H}}_{2}\\text{O}[\/latex]<\/p>\r\n\r\n<table id=\"fs-idp140513680\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and four rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters C, H, and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 1 equals 1,\u201d \u201c4 times 1 equals 4,\u201d and \u201c2 times 2 equals 4.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 1 equals 1,\u201d \u201c2 times 2 equals 4,\u201d and \u201c( 1 times 2 ) plus ( 2 times 1 ) equals 4.\u201d Under the \u201cBalanced?\u201d column are, \u201c1 equals 1, yes,\u201d \u201c4 equals 4, yes,\u201d \u201c4 equals 4, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>C<\/td>\r\n<td>1 \u00d7 1 = 1<\/td>\r\n<td>1 \u00d7 1 = 1<\/td>\r\n<td>1 = 1, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td>4 \u00d7 1 = 4<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td>(1 \u00d7 2) + (2 \u00d7 1) = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nA balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an <em>unbalanced<\/em> chemical equation:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{H}}_{2}\\text{O}\\rightarrow{\\text{H}}_{2}+{\\text{O}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\r\nComparing the number of H and O atoms on either side of this equation confirms its imbalance:\r\n<table id=\"fs-idp104786160\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 1 equals 1.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c1 does not equal 2, no.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>1 \u00d7 1 = 1<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u2260 2, no<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the <em>coefficients<\/em> of the equation may be changed as needed. Keep in mind, of course, that the <em>formula subscripts<\/em> define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H<sub>2<\/sub>O to H<sub>2<\/sub>O<sub>2<\/sub> would yield balance in the number of atoms, but doing so also changes the reactant\u2019s identity (it\u2019s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H<sub>2<\/sub>O to 2.\r\n<p style=\"text-align: center\">[latex]\\large\\mathbf{2}\\text{ H}_{2}\\text{O}\\rightarrow{\\text{H}}_{2}+{\\text{O}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\r\n\r\n<table id=\"fs-idm15543696\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 4\u201d equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 does not equal 2, no,\u201d and \u201c2 equals 2, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>4 \u2260 2, no<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>2 \u00d7 1 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H<sub>2<\/sub> product to 2.\r\n<p style=\"text-align: center\">[latex]\\large2{\\text{ H}}_{2}\\text{O}\\rightarrow\\mathbf{2}{\\text{ H}}_{2}+{\\text{O}}_{2}\\text{ (balanced)}[\/latex]<\/p>\r\n\r\n<table id=\"fs-idp151419504\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 2\u201d equation is bold. Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c2 equals 2, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 2<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>2 \u00d7 1 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThese coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:\r\n<p style=\"text-align: center\">[latex]\\large2{\\text{ H}}_{2}\\text{O}\\rightarrow 2{\\text{ H}}_{2}+{\\text{O}}_{2}[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1: <strong>Balancing Chemical Equations<\/strong><\/h3>\r\nWrite a balanced equation for the reaction of molecular nitrogen (N<sub>2<\/sub>) and oxygen (O<sub>2<\/sub>) to form dinitrogen pentoxide.\r\n\r\n[reveal-answer q=\"463373\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"463373\"]\r\n\r\nFirst, write the unbalanced equation:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{N}}_{2}+{\\text{O}}_{2}\\rightarrow{\\text{N}}_{2}{\\text{O}}_{5}\\text{ (unbalanced)}[\/latex]<\/p>\r\nNext, count the number of each type of atom present in the unbalanced equation.\r\n<table id=\"fs-idp107503280\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 5 equals 5.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c2 does not equal 5, no.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>N<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u00d7 5 = 5<\/td>\r\n<td>2 \u2260 5, no<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThough nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O<sub>2<\/sub> and N<sub>2<\/sub>O<sub>5<\/sub> to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).\r\n<p style=\"text-align: center\">[latex]\\large{\\text{N}}_{2}+\\mathbf{5}{\\text{ O}}_{2}\\rightarrow\\mathbf{2}{\\text{ N}}_{2}{\\text{O}}_{5}\\text{ (unbalanced)}[\/latex]<\/p>\r\n\r\n<table id=\"fs-idp7305424\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c5 times 2 equals 10.\u201d The 5 in the second equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d The initial 2 in each equation is bold. Under the \u201cBalanced?\u201d column are, \u201c2 does not equal 4, no,\u201d and \u201c10 equals 10, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>N<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td>2 \u2260 4, no<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td><strong>5<\/strong> \u00d7 2 = 10<\/td>\r\n<td><strong>2<\/strong> \u00d7 5 = 10<\/td>\r\n<td>10 = 10, yes<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N<sub>2<\/sub> to 2.\r\n<p style=\"text-align: center\">[latex]\\large2{\\text{ N}}_{2}+5{\\text{ O}}_{2}\\rightarrow 2{\\text{ N}}_{2}{\\text{O}}_{5}[\/latex]<\/p>\r\n\r\n<table id=\"fs-idm9607408\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c5 times 2 equals 10.\u201d The initial 2 in the first equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c10 equals 10, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>N<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>5 \u00d7 2 = 10<\/td>\r\n<td>2 \u00d7 5 = 10<\/td>\r\n<td>10 = 10, yes<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWrite a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)\r\n\r\n[reveal-answer q=\"267993\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"267993\"][latex]\\large2{\\text{ NH}}_{4}{\\text{NO}}_{3}\\rightarrow 2{\\text{ N}}_{2}+{\\text{O}}_{2}+4{\\text{ H}}_{2}\\text{O}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIt is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation\u2019s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C<sub>2<\/sub>H<sub>6<\/sub>) with oxygen to yield H<sub>2<\/sub>O and CO<sub>2<\/sub>, represented by the unbalanced equation:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{C}}_{2}{\\text{H}}_{6}+{\\text{O}}_{2}\\rightarrow{\\text{H}}_{2}\\text{O}+{\\text{CO}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\r\nFollowing the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{C}}_{2}{\\text{H}}_{6}+{\\text{O}}_{2}\\rightarrow 3{\\text{ H}}_{2}\\text{O}+2{\\text{ CO}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\r\nThis results in seven O atoms on the product side of the equation, an odd number\u2014no integer coefficient can be used with the O<sub>2<\/sub> reactant to yield an odd number, so a fractional coefficient, [latex]\\displaystyle\\frac{7}{2}[\/latex] , is used instead to yield a provisional balanced equation:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{C}}_{2}{\\text{H}}_{6}+\\frac{7}{2}{\\text{ O}}_{2}\\rightarrow 3{\\text{ H}}_{2}\\text{O}+2{\\text{ CO}}_{2}[\/latex]<\/p>\r\nA conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:\r\n<p style=\"text-align: center\">[latex]\\large2{\\text{ C}}_{2}{\\text{H}}_{6}+7{\\text{ O}}_{2}\\rightarrow 6{\\text{ H}}_{2}\\text{O}+4{\\text{ CO}}_{2}[\/latex]<\/p>\r\nFinally with regard to balanced equations, recall that convention dictates use of the <em>smallest whole-number coefficients<\/em>. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,\r\n<p style=\"text-align: center\">[latex]\\large3{\\text{ N}}_{2}+9{\\text{ H}}_{2}\\rightarrow 6{\\text{ NH}}_{3}[\/latex]<\/p>\r\nthe coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:\r\n<p style=\"text-align: center\">[latex]\\large{\\text{N}}_{2}+3{\\text{ H}}_{2}\\rightarrow 2{\\text{ NH}}_{3}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">Use <a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/balancing-chemical-equations\" target=\"_blank\" rel=\"noopener noreferrer\">this interactive PhET tutorial for additional practice balancing equations<\/a>.<\/div>\r\n<h2>Additional Information in Chemical Equations<\/h2>\r\nThe physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include <em>s<\/em> for solids, <em>l<\/em> for liquids, <em>g<\/em> for gases, and <em>aq<\/em> for substances dissolved in water (<em>aqueous solutions<\/em>, as introduced in the preceding chapter). These notations are illustrated in the example equation here:\r\n<p style=\"text-align: center\">[latex]\\large2\\text{ Na(}s\\text{)}+2{\\text{ H}}_{2}\\text{O(}l\\text{)}\\rightarrow 2\\text{ NaOH(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\r\nThis equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).\r\n\r\nSpecial conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation\u2019s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (\u0394) over the arrow.\r\n<p style=\"text-align: center\">[latex]\\large{\\text{CaCO}}_{3}\\text{(}s\\text{)}\\stackrel{\\Delta}{\\rightarrow}\\text{CaO(}s\\text{)}+{\\text{CO}}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\r\nOther examples of these special conditions will be encountered in more depth in later chapters.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nChemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li style=\"list-style-type: none\">\r\n<ol>\r\n \t<li>What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?<\/li>\r\n \t<li>Balance the following equations:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]\\large\\text{Ag}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{S}\\text{(}g\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{Ag}}_{2}\\text{S}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]{\\large\\text{P}}_{4}\\text{(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{P}}_{4}{\\text{O}}_{10}\\text{(}s\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\text{Pb(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{Pb(OH)}}_{2}\\text{(}s\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\text{Fe(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow{\\text{Fe}}_{3}{\\text{O}}_{4}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{Sc}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}+{\\text{SO}}_{3}\\text{(}l\\text{)}\\rightarrow{\\text{Sc}}_{2}{\\text{(}{\\text{SO}}_{4}\\text{)}}_{3}\\text{(}s\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{Ca}}_{3}{\\text{(}{\\text{PO}}_{4}\\text{)}}_{2}\\text{(}aq\\text{)}+{\\text{H}}_{3}{\\text{PO}}_{4}\\text{(}aq\\text{)}\\rightarrow\\text{Ca}{\\text{(}{\\text{H}}_{2}{\\text{PO}}_{4}\\text{)}}_{2}\\text{(}aq\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\text{Al(}s\\text{)}+{\\text{H}}_{2}{\\text{SO}}_{4}\\text{(}aq\\text{)}\\rightarrow{\\text{Al}}_{2}{\\text{(}{\\text{SO}}_{4}\\text{)}}_{3}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{TiCl}}_{4}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}\\rightarrow{\\text{TiO}}_{2}\\text{(}s\\text{)}+\\text{HCl(}g\\text{)}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Balance the following equations:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]\\large{\\text{PCl}}_{5}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow{\\text{POCl}}_{3}\\text{(}l\\text{)}+\\text{HCl(}aq\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\text{Cu(}s\\text{)}+{\\text{HNO}}_{3}\\text{(}aq\\text{)}\\rightarrow\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}+\\text{NO(}g\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{H}}_{2}\\text{(}g\\text{)}+{\\text{I}}_{2}\\text{(}s\\text{)}\\rightarrow\\text{HI(}s\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\text{Fe(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{Fe}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\text{Na(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow\\text{NaOH}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{Cr}}_{2}{\\text{O}}_{7}\\text{(}s\\text{)}\\rightarrow{\\text{Cr}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}+{\\text{N}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{P}}_{4}\\text{(}s\\text{)}+{\\text{Cl}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{PCl}}_{3}\\text{(}l\\text{)}[\/latex]<\/li>\r\n \t<li>[latex]\\large{\\text{PtCl}}_{4}\\text{(}s\\text{)}\\rightarrow\\text{Pt}\\text{(}s\\text{)}+{\\text{Cl}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"812718\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"812718\"]\r\n\r\n1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.\r\n\r\n3.\r\n\r\n(a) [latex]\\large{\\text{PCl}}_{5}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow{\\text{POCl}}_{3}\\text{(}l\\text{)}+2\\text{ HCl(}aq\\text{);}[\/latex]\r\n\r\n(b) [latex]\\large3\\text{ Cu}\\text{(}s\\text{)}+8{\\text{ HNO}}_{3}\\text{(}aq\\text{)}\\rightarrow 3\\text{ Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+4{\\text{ H}}_{2}\\text{O(}l\\text{)}+2\\text{ NO(}g\\text{);}[\/latex]\r\n\r\n(c) [latex]\\large{\\text{H}}_{2}\\text{(}g\\text{)}+{\\text{I}}_{2}\\text{(}s\\text{)}\\rightarrow 2\\text{ HI(}s\\text{);}[\/latex]\r\n\r\n(d) [latex]\\large4\\text{ Fe(}s\\text{)}+3{\\text{ O}}_{2}\\text{(}g\\text{)}\\rightarrow 2{\\text{ Fe}}_{2}{\\text{O}}_{3}\\text{(}s\\text{);}[\/latex]\r\n\r\n(e) [latex]\\large2\\text{ Na(}s\\text{)}+2{\\text{ H}}_{2}\\text{O(}l\\text{)}\\rightarrow 2\\text{ NaOH(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{);}[\/latex]\r\n\r\n(f) [latex]\\large{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}+{\\text{Cr}}_{5}{\\text{ O}}_{7}\\text{(}s\\text{)}\\rightarrow{\\text{Cr}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}+{\\text{N}}_{2}\\text{(}g\\text{)}+4{\\text{ H}}_{2}\\text{O(}g\\text{);}[\/latex]\r\n\r\n(g) [latex]\\large{\\text{P}}_{4}\\text{(}s\\text{)}+6{\\text{ Cl}}_{2}\\text{(}g\\text{)}\\rightarrow 4{\\text{ PCl}}_{3}\\text{(}l\\text{);}[\/latex]\r\n\r\n(h) [latex]\\large{\\text{PtCl}}_{4}\\text{(}s\\text{)}\\rightarrow\\text{Pt(}s\\text{)}+2{\\text{ Cl}}_{2}\\text{(}g\\text{)}[\/latex]\r\n\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>balanced equation: <\/strong>chemical equation with equal numbers of atoms for each element in the reactant and product\r\n\r\n<strong>chemical equation: <\/strong>symbolic representation of a chemical reaction\r\n\r\n<strong>coefficient: <\/strong>number placed in front of symbols or formulas in a chemical equation to indicate their relative amount\r\n\r\n<strong>product: <\/strong>substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation\r\n\r\n<strong>reactant: <\/strong>substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Derive chemical equations from narrative descriptions of chemical reactions.<\/li>\n<li>Write and balance chemical equations.<\/li>\n<\/ul>\n<\/div>\n<p>When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a <strong>chemical equation<\/strong>. Consider as an example the reaction between one methane molecule (CH<sub>4<\/sub>) and two diatomic oxygen molecules (O<sub>2<\/sub>) to produce one carbon dioxide molecule (CO<sub>2<\/sub>) and two water molecules (H<sub>2<\/sub>O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.<\/p>\n<div style=\"width: 891px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211218\/CNX_Chem_04_01_rxn21.jpg\" alt=\"This figure shows a balanced chemical equation followed below by a representation of the equation using space-filling models. The equation reads C H subscript 4 plus 2 O subscript 2 arrow C O subscript 2 plus 2 H subscript 2 O. Under the C H subscript 4, the molecule is shown with a central black sphere, representing a C atom, to which 4 smaller white spheres, representing H atoms, are distributed evenly around. All four H atoms are bonded to the central black C atom. This is followed by a plus sign. Under the 2 O subscript 2, two molecules are shown. The molecules are each composed of two red spheres bonded together. The red spheres represent O atoms. To the right of an arrow and under the C O subscript 2, appears a single molecule with a black central sphere with two red spheres bonded to the left and right. Following a plus sign and under the 2 H subscript 2 O, are two molecules, each with a central red sphere and two smaller white spheres attached to the lower right and lower left sides of the central red sphere. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms.\" width=\"881\" height=\"401\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).<\/p>\n<\/div>\n<p>This example illustrates the fundamental aspects of any chemical equation:<\/p>\n<ol>\n<li>The substances undergoing reaction are called <strong>reactants<\/strong>, and their formulas are placed on the left side of the equation.<\/li>\n<li>The substances generated by the reaction are called <strong>products<\/strong>, and their formulas are placed on the right sight of the equation.<\/li>\n<li>Plus signs (+) separate individual reactant and product formulas, and an arrow ([latex]\\rightarrow[\/latex]) separates the reactant and product (left and right) sides of the equation.<\/li>\n<li>The relative numbers of reactant and product species are represented by <strong>coefficients<\/strong> (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.<\/li>\n<\/ol>\n<p>It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the <em>relative<\/em> numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 2). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:<\/p>\n<ul id=\"fs-idp42981680\">\n<li><em>One<\/em> methane molecule and <em>two<\/em> oxygen molecules react to yield <em>one<\/em> carbon dioxide molecule and <em>two<\/em> water molecules.<\/li>\n<li><em>One dozen<\/em> methane molecules and <em>two dozen<\/em> oxygen molecules react to yield <em>one dozen<\/em> carbon dioxide molecules and <em>two dozen<\/em> water molecules.<\/li>\n<li><em>One mole<\/em> of methane molecules and <em>2 moles<\/em> of oxygen molecules react to yield <em>1 mole<\/em> of carbon dioxide molecules and <em>2 moles<\/em> of water molecules.<\/li>\n<\/ul>\n<div style=\"width: 890px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211219\/CNX_Chem_04_01_rxn31.jpg\" alt=\"This image has a left side, labeled, \u201cMixture before reaction\u201d separated by a vertical dashed line from right side labeled, \u201cMixture after reaction.\u201d On the left side of the figure, two types of molecules are illustrated with space-filling models. Six of the molecules have only two red spheres bonded together. Three of the molecules have four small white spheres evenly distributed about and bonded to a central, larger black sphere. On the right side of the dashed vertical line, two types of molecules which are different from those on the left side are shown. Six of the molecules have a central red sphere to which smaller white spheres are bonded. The white spheres are not opposite each other on the red atoms, giving the molecule a bent shape or appearance. The second molecule type has a central black sphere to which two red spheres are attached on opposite sides, resulting in a linear shape or appearance. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms. On each side of the dashed line, twelve red, three black, and twelve white spheres are present.\" width=\"880\" height=\"341\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Regardless of the absolute number of molecules involved, the ratios between numbers of molecules are the same as that given in the chemical equation.<\/p>\n<\/div>\n<h2>Balancing Equations<\/h2>\n<p>A <strong>balanced chemical is equation <\/strong>has equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element\u2019s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO<sub>2<\/sub> and H<sub>2<\/sub>O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is<\/p>\n<p style=\"text-align: center\">[latex]\\large\\left(1{\\text{ CO}}_{2}\\text{ molecule }\\times \\frac{\\text{2 O atoms}}{{\\text{1 CO}}_{2}\\text{ molecule }}\\right)+\\left(2{\\text{ H}}_{2}\\text{O molecule }\\times \\frac{\\text{1 O atom}}{{\\text{1 H}}_{2}\\text{O molecule }}\\right)=\\text{4 O atoms}[\/latex]<\/p>\n<p>The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{CH}}_{4}+2{\\text{ O}}_{2}\\rightarrow{\\text{CO}}_{2}+2{\\text{ H}}_{2}\\text{O}[\/latex]<\/p>\n<table id=\"fs-idp140513680\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and four rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters C, H, and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 1 equals 1,\u201d \u201c4 times 1 equals 4,\u201d and \u201c2 times 2 equals 4.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 1 equals 1,\u201d \u201c2 times 2 equals 4,\u201d and \u201c( 1 times 2 ) plus ( 2 times 1 ) equals 4.\u201d Under the \u201cBalanced?\u201d column are, \u201c1 equals 1, yes,\u201d \u201c4 equals 4, yes,\u201d \u201c4 equals 4, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>C<\/td>\n<td>1 \u00d7 1 = 1<\/td>\n<td>1 \u00d7 1 = 1<\/td>\n<td>1 = 1, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<\/td>\n<td>4 \u00d7 1 = 4<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td>(1 \u00d7 2) + (2 \u00d7 1) = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an <em>unbalanced<\/em> chemical equation:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{H}}_{2}\\text{O}\\rightarrow{\\text{H}}_{2}+{\\text{O}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\n<p>Comparing the number of H and O atoms on either side of this equation confirms its imbalance:<\/p>\n<table id=\"fs-idp104786160\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 1 equals 1.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c1 does not equal 2, no.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>H<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>1 \u00d7 1 = 1<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u2260 2, no<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the <em>coefficients<\/em> of the equation may be changed as needed. Keep in mind, of course, that the <em>formula subscripts<\/em> define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H<sub>2<\/sub>O to H<sub>2<\/sub>O<sub>2<\/sub> would yield balance in the number of atoms, but doing so also changes the reactant\u2019s identity (it\u2019s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H<sub>2<\/sub>O to 2.<\/p>\n<p style=\"text-align: center\">[latex]\\large\\mathbf{2}\\text{ H}_{2}\\text{O}\\rightarrow{\\text{H}}_{2}+{\\text{O}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\n<table id=\"fs-idm15543696\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 4\u201d equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 does not equal 2, no,\u201d and \u201c2 equals 2, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>H<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>4 \u2260 2, no<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>2 \u00d7 1 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H<sub>2<\/sub> product to 2.<\/p>\n<p style=\"text-align: center\">[latex]\\large2{\\text{ H}}_{2}\\text{O}\\rightarrow\\mathbf{2}{\\text{ H}}_{2}+{\\text{O}}_{2}\\text{ (balanced)}[\/latex]<\/p>\n<table id=\"fs-idp151419504\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 2\u201d equation is bold. Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c2 equals 2, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>H<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 2<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>2 \u00d7 1 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:<\/p>\n<p style=\"text-align: center\">[latex]\\large2{\\text{ H}}_{2}\\text{O}\\rightarrow 2{\\text{ H}}_{2}+{\\text{O}}_{2}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1: <strong>Balancing Chemical Equations<\/strong><\/h3>\n<p>Write a balanced equation for the reaction of molecular nitrogen (N<sub>2<\/sub>) and oxygen (O<sub>2<\/sub>) to form dinitrogen pentoxide.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q463373\">Show Answer<\/span><\/p>\n<div id=\"q463373\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, write the unbalanced equation:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{N}}_{2}+{\\text{O}}_{2}\\rightarrow{\\text{N}}_{2}{\\text{O}}_{5}\\text{ (unbalanced)}[\/latex]<\/p>\n<p>Next, count the number of each type of atom present in the unbalanced equation.<\/p>\n<table id=\"fs-idp107503280\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 5 equals 5.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c2 does not equal 5, no.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>N<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u00d7 5 = 5<\/td>\n<td>2 \u2260 5, no<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O<sub>2<\/sub> and N<sub>2<\/sub>O<sub>5<\/sub> to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{N}}_{2}+\\mathbf{5}{\\text{ O}}_{2}\\rightarrow\\mathbf{2}{\\text{ N}}_{2}{\\text{O}}_{5}\\text{ (unbalanced)}[\/latex]<\/p>\n<table id=\"fs-idp7305424\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c5 times 2 equals 10.\u201d The 5 in the second equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d The initial 2 in each equation is bold. Under the \u201cBalanced?\u201d column are, \u201c2 does not equal 4, no,\u201d and \u201c10 equals 10, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>N<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td>2 \u2260 4, no<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td><strong>5<\/strong> \u00d7 2 = 10<\/td>\n<td><strong>2<\/strong> \u00d7 5 = 10<\/td>\n<td>10 = 10, yes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N<sub>2<\/sub> to 2.<\/p>\n<p style=\"text-align: center\">[latex]\\large2{\\text{ N}}_{2}+5{\\text{ O}}_{2}\\rightarrow 2{\\text{ N}}_{2}{\\text{O}}_{5}[\/latex]<\/p>\n<table id=\"fs-idm9607408\" class=\"medium-unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c5 times 2 equals 10.\u201d The initial 2 in the first equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c10 equals 10, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>N<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>5 \u00d7 2 = 10<\/td>\n<td>2 \u00d7 5 = 10<\/td>\n<td>10 = 10, yes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q267993\">Show Answer<\/span><\/p>\n<div id=\"q267993\" class=\"hidden-answer\" style=\"display: none\">[latex]\\large2{\\text{ NH}}_{4}{\\text{NO}}_{3}\\rightarrow 2{\\text{ N}}_{2}+{\\text{O}}_{2}+4{\\text{ H}}_{2}\\text{O}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation\u2019s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C<sub>2<\/sub>H<sub>6<\/sub>) with oxygen to yield H<sub>2<\/sub>O and CO<sub>2<\/sub>, represented by the unbalanced equation:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{C}}_{2}{\\text{H}}_{6}+{\\text{O}}_{2}\\rightarrow{\\text{H}}_{2}\\text{O}+{\\text{CO}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\n<p>Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{C}}_{2}{\\text{H}}_{6}+{\\text{O}}_{2}\\rightarrow 3{\\text{ H}}_{2}\\text{O}+2{\\text{ CO}}_{2}\\text{ (unbalanced)}[\/latex]<\/p>\n<p>This results in seven O atoms on the product side of the equation, an odd number\u2014no integer coefficient can be used with the O<sub>2<\/sub> reactant to yield an odd number, so a fractional coefficient, [latex]\\displaystyle\\frac{7}{2}[\/latex] , is used instead to yield a provisional balanced equation:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{C}}_{2}{\\text{H}}_{6}+\\frac{7}{2}{\\text{ O}}_{2}\\rightarrow 3{\\text{ H}}_{2}\\text{O}+2{\\text{ CO}}_{2}[\/latex]<\/p>\n<p>A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:<\/p>\n<p style=\"text-align: center\">[latex]\\large2{\\text{ C}}_{2}{\\text{H}}_{6}+7{\\text{ O}}_{2}\\rightarrow 6{\\text{ H}}_{2}\\text{O}+4{\\text{ CO}}_{2}[\/latex]<\/p>\n<p>Finally with regard to balanced equations, recall that convention dictates use of the <em>smallest whole-number coefficients<\/em>. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,<\/p>\n<p style=\"text-align: center\">[latex]\\large3{\\text{ N}}_{2}+9{\\text{ H}}_{2}\\rightarrow 6{\\text{ NH}}_{3}[\/latex]<\/p>\n<p>the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{N}}_{2}+3{\\text{ H}}_{2}\\rightarrow 2{\\text{ NH}}_{3}[\/latex]<\/p>\n<div class=\"textbox\">Use <a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/balancing-chemical-equations\" target=\"_blank\" rel=\"noopener noreferrer\">this interactive PhET tutorial for additional practice balancing equations<\/a>.<\/div>\n<h2>Additional Information in Chemical Equations<\/h2>\n<p>The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include <em>s<\/em> for solids, <em>l<\/em> for liquids, <em>g<\/em> for gases, and <em>aq<\/em> for substances dissolved in water (<em>aqueous solutions<\/em>, as introduced in the preceding chapter). These notations are illustrated in the example equation here:<\/p>\n<p style=\"text-align: center\">[latex]\\large2\\text{ Na(}s\\text{)}+2{\\text{ H}}_{2}\\text{O(}l\\text{)}\\rightarrow 2\\text{ NaOH(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<p>This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).<\/p>\n<p>Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation\u2019s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (\u0394) over the arrow.<\/p>\n<p style=\"text-align: center\">[latex]\\large{\\text{CaCO}}_{3}\\text{(}s\\text{)}\\stackrel{\\Delta}{\\rightarrow}\\text{CaO(}s\\text{)}+{\\text{CO}}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<p>Other examples of these special conditions will be encountered in more depth in later chapters.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li style=\"list-style-type: none\">\n<ol>\n<li>What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?<\/li>\n<li>Balance the following equations:\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]\\large\\text{Ag}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{S}\\text{(}g\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{Ag}}_{2}\\text{S}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\n<li>[latex]{\\large\\text{P}}_{4}\\text{(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{P}}_{4}{\\text{O}}_{10}\\text{(}s\\text{)}[\/latex]<\/li>\n<li>[latex]\\large\\text{Pb(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{Pb(OH)}}_{2}\\text{(}s\\text{)}[\/latex]<\/li>\n<li>[latex]\\large\\text{Fe(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow{\\text{Fe}}_{3}{\\text{O}}_{4}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{Sc}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}+{\\text{SO}}_{3}\\text{(}l\\text{)}\\rightarrow{\\text{Sc}}_{2}{\\text{(}{\\text{SO}}_{4}\\text{)}}_{3}\\text{(}s\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{Ca}}_{3}{\\text{(}{\\text{PO}}_{4}\\text{)}}_{2}\\text{(}aq\\text{)}+{\\text{H}}_{3}{\\text{PO}}_{4}\\text{(}aq\\text{)}\\rightarrow\\text{Ca}{\\text{(}{\\text{H}}_{2}{\\text{PO}}_{4}\\text{)}}_{2}\\text{(}aq\\text{)}[\/latex]<\/li>\n<li>[latex]\\large\\text{Al(}s\\text{)}+{\\text{H}}_{2}{\\text{SO}}_{4}\\text{(}aq\\text{)}\\rightarrow{\\text{Al}}_{2}{\\text{(}{\\text{SO}}_{4}\\text{)}}_{3}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{TiCl}}_{4}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}\\rightarrow{\\text{TiO}}_{2}\\text{(}s\\text{)}+\\text{HCl(}g\\text{)}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Balance the following equations:\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]\\large{\\text{PCl}}_{5}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow{\\text{POCl}}_{3}\\text{(}l\\text{)}+\\text{HCl(}aq\\text{)}[\/latex]<\/li>\n<li>[latex]\\large\\text{Cu(}s\\text{)}+{\\text{HNO}}_{3}\\text{(}aq\\text{)}\\rightarrow\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}+\\text{NO(}g\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{H}}_{2}\\text{(}g\\text{)}+{\\text{I}}_{2}\\text{(}s\\text{)}\\rightarrow\\text{HI(}s\\text{)}[\/latex]<\/li>\n<li>[latex]\\large\\text{Fe(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{Fe}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}[\/latex]<\/li>\n<li>[latex]\\large\\text{Na(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow\\text{NaOH}\\text{(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{Cr}}_{2}{\\text{O}}_{7}\\text{(}s\\text{)}\\rightarrow{\\text{Cr}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}+{\\text{N}}_{2}\\text{(}g\\text{)}+{\\text{H}}_{2}\\text{O(}g\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{P}}_{4}\\text{(}s\\text{)}+{\\text{Cl}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{PCl}}_{3}\\text{(}l\\text{)}[\/latex]<\/li>\n<li>[latex]\\large{\\text{PtCl}}_{4}\\text{(}s\\text{)}\\rightarrow\\text{Pt}\\text{(}s\\text{)}+{\\text{Cl}}_{2}\\text{(}g\\text{)}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812718\">Show Selected Answers<\/span><\/p>\n<div id=\"q812718\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.<\/p>\n<p>3.<\/p>\n<p>(a) [latex]\\large{\\text{PCl}}_{5}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O(}l\\text{)}\\rightarrow{\\text{POCl}}_{3}\\text{(}l\\text{)}+2\\text{ HCl(}aq\\text{);}[\/latex]<\/p>\n<p>(b) [latex]\\large3\\text{ Cu}\\text{(}s\\text{)}+8{\\text{ HNO}}_{3}\\text{(}aq\\text{)}\\rightarrow 3\\text{ Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}+4{\\text{ H}}_{2}\\text{O(}l\\text{)}+2\\text{ NO(}g\\text{);}[\/latex]<\/p>\n<p>(c) [latex]\\large{\\text{H}}_{2}\\text{(}g\\text{)}+{\\text{I}}_{2}\\text{(}s\\text{)}\\rightarrow 2\\text{ HI(}s\\text{);}[\/latex]<\/p>\n<p>(d) [latex]\\large4\\text{ Fe(}s\\text{)}+3{\\text{ O}}_{2}\\text{(}g\\text{)}\\rightarrow 2{\\text{ Fe}}_{2}{\\text{O}}_{3}\\text{(}s\\text{);}[\/latex]<\/p>\n<p>(e) [latex]\\large2\\text{ Na(}s\\text{)}+2{\\text{ H}}_{2}\\text{O(}l\\text{)}\\rightarrow 2\\text{ NaOH(}aq\\text{)}+{\\text{H}}_{2}\\text{(}g\\text{);}[\/latex]<\/p>\n<p>(f) [latex]\\large{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}+{\\text{Cr}}_{5}{\\text{ O}}_{7}\\text{(}s\\text{)}\\rightarrow{\\text{Cr}}_{2}{\\text{O}}_{3}\\text{(}s\\text{)}+{\\text{N}}_{2}\\text{(}g\\text{)}+4{\\text{ H}}_{2}\\text{O(}g\\text{);}[\/latex]<\/p>\n<p>(g) [latex]\\large{\\text{P}}_{4}\\text{(}s\\text{)}+6{\\text{ Cl}}_{2}\\text{(}g\\text{)}\\rightarrow 4{\\text{ PCl}}_{3}\\text{(}l\\text{);}[\/latex]<\/p>\n<p>(h) [latex]\\large{\\text{PtCl}}_{4}\\text{(}s\\text{)}\\rightarrow\\text{Pt(}s\\text{)}+2{\\text{ Cl}}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>balanced equation: <\/strong>chemical equation with equal numbers of atoms for each element in the reactant and product<\/p>\n<p><strong>chemical equation: <\/strong>symbolic representation of a chemical reaction<\/p>\n<p><strong>coefficient: <\/strong>number placed in front of symbols or formulas in a chemical equation to indicate their relative amount<\/p>\n<p><strong>product: <\/strong>substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation<\/p>\n<p><strong>reactant: <\/strong>substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation<\/p>\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><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/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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