4.24: Balancing Chemical Equations: Additional Examples Last updated Save as PDF Page ID218409 \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\) Learning Objectives Use coefficients to balance a chemical equation. Example \(\PageIndex{1}\) Balance the following chemical equation by writing coefficients in the "blanks," as necessary. ___ \(\ce{Mg(OH)_2} \left( aq \right) +\) ___ \(\ce{(NH_4)_2SO_4} \left( aq \right) \rightarrow\) ___ \(\ce{NH_4OH} \left( aq \right) +\) ___ \(\ce{MgSO_4} \left( s \right)\) Solution In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. The first reactant in this equation contains magnesium, Mg, and a hydroxide ion, OH-1, which is a polyatomic anion. The second reactant is comprised of both a polyatomic cation, the ammonium ion, NH4+1, and a polyatomic anion, the sulfate ion, SO4-2. Because polyatomic ions contain multiple atoms and generally react as indivisible units, any polyatomic ion that is present in both a reactant and a product in a given chemical equation should be treated as a singular entity. Therefore, since each of the polyatomic ions that is listed above is also contained in one of the chemicals shown on the product side of the reaction arrow, each of these ions should be balanced as a single entity. Finally, recall that parentheses are utilized in ionic chemical formulas to offset polyatomic ions as units, and the subscript that is written outside of the parentheses indicates the how many of that ion are present within the ionic compound. If parentheses are not explicitly-written around the formula of a polyatomic ion, an unwritten subscript of "1" is understood to correspond to that polyatomic unit. The quantities in which these chemicals are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced Mg 1 1 OH-1 2 1 NH4+1 2 1 SO4-2 1 1 In order for a component of a reaction to be balanced, the reactants and products of the reaction must contain identical total quantities of that type of element or polyatomic ion. Since both sides of the reaction contain equal amounts of magnesium, Mg, and sulfate ions, SO4-2, these components are balanced. However, the hydroxide ion, OH-1, and the ammonium ion, NH4+1 are not balanced, because each of these polyatomic ions is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. Neither of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow or is represented in its atomic form. Therefore, the balancing process could be initiated by selecting either the hydroxide ion, OH-1, or the ammonium ion, NH4+1, for further analysis. In order to balance the hydroxide ion, OH-1, a coefficient should be written in the "blank" that corresponds to ammonium hydroxide, NH4OH, on the right side of the equation, as fewer hydroxide ions are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{Mg(OH)_2} \left( aq \right) +\) ___ \(\ce{(NH_4)_2SO_4} \left( aq \right) \rightarrow\) 2 \(\ce{NH_4OH} \left( aq \right) +\) ___ \(\ce{MgSO_4} \left( s \right)\) As ammonium hydroxide, NH4OH, contains both the ammonium ion, NH4+1, and the hydroxide ion, OH-1, incorporating this coefficient alters the amounts of both of these polyatomic ions on the product side of the equation. The updated quantities of these ions, which are obtained by multiplying the subscripts that are associated with each ion symbol by the coefficient that was written, are reflected in the table that is shown below. Inserting this coefficient balances the hydroxide ion, OH-1, as intended. Additionally, the quantities in which the ammonium ion, NH4+1, are present are now also identical, and this ion is now balanced, as well. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced Mg 1 1 OH-1 2 \( \cancel{\rm{1} } \) 2 NH4+1 2 \( \cancel{\rm{1} } \) 2 SO4-2 1 1 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1"s that are understood to occupy the first, second, and last "blanks" in this equation, this coefficient cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Exercise \(\PageIndex{1}\) Balance the following chemical equation by writing coefficients in the "blanks," as necessary. ___ \(\ce{Ag_2S} \left( s \right) +\) ___ \(\ce{Al} \left( s \right) \rightarrow\) ___ \(\ce{Ag} \left( s \right) +\) ___ \(\ce{Al_2S_3} \left( s \right)\) Answer In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of silver, Ag, sulfur, S, and aluminum, Al, that are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced Ag 2 1 S 1 3 Al 1 2 In order for a component of a reaction to be balanced, the reactants and products of the reaction must contain identical total quantities of that type of element or polyatomic ion. Therefore, none of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. None of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow. However, as both silver, Ag, and aluminum, Al, are represented in their atomic forms, the balancing process should not be initiated using either of these elements. Therefore, sulfur, S, should be the first element that is balanced in this equation. In order to balance sulfur, S, a coefficient should be written in the "blank" that corresponds to silver sulfide, Ag2S, on the left side of the equation, as fewer sulfurs are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 3, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 3 \(\ce{Ag_2S} \left( s \right) +\) ___ \(\ce{Al} \left( s \right) \rightarrow\) ___ \(\ce{Ag} \left( s \right) +\) ___ \(\ce{Al_2S_3} \left( s \right)\) As silver sulfide, Ag2S, contains both silver, Ag, and sulfur, S, incorporating this coefficient alters the amounts of both of these elements on the reactant side of the equation. The updated quantities of these elements, which are obtained by multiplying the subscripts that are associated with each elemental symbol by the coefficient that was written, are reflected in the table that is shown below. Inserting this coefficient balances sulfur, S, as intended. The quantities in which silver, Ag, and aluminum, Al, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Ag \( \cancel{\rm{2} } \) 6 1 S \( \cancel{\rm{1} } \) 3 3 Al 1 2 Because both silver, Ag, and aluminum, Al, are represented in their atomic forms, the balancing process could be continued by selecting either of these elements for further analysis. In order to balance silver, Ag, a coefficient should be written in the "blank" that corresponds to this element on the right side of the equation, as fewer silvers are present on this side of the reaction arrow. The value of this coefficient, 6, is determined by dividing the larger quantity of this element, 6, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 3 \(\ce{Ag_2S} \left( s \right) +\) ___ \(\ce{Al} \left( s \right) \rightarrow\) 6 \(\ce{Ag} \left( s \right) +\) ___ \(\ce{Al_2S_3} \left( s \right)\) As silver, Ag, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the product side of the equation. The updated quantity of this element, which was obtained by multiplying its subscript by the coefficient that was written, is reflected in the table that is shown below. Inserting this coefficient balances silver, Ag, as intended. The quantities in which aluminum, Al, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Ag \( \cancel{\rm{2} } \) 6 \( \bcancel{\rm{1} } \) 6 S \( \cancel{\rm{1} } \) 3 3 Al 1 2 In order to balance aluminum, Al, a coefficient should be written in the "blank" that corresponds to this element on the left side of the equation, as fewer aluminums are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 3 \(\ce{Ag_2S} \left( s \right) +\) 2 \(\ce{Al} \left( s \right) \rightarrow\) 6 \(\ce{Ag} \left( s \right) +\) ___ \(\ce{Al_2S_3} \left( s \right)\) As aluminum, Al, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the reactant side of the equation. The updated quantity of this element, which was obtained by multiplying its subscript by the coefficient that was written, is reflected in the table that is shown below. Inserting this coefficient balances aluminum, Al, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced Ag \( \cancel{\rm{2} } \) 6 \( \bcancel{\rm{1} } \) 6 S \( \cancel{\rm{1} } \) 3 3 Al \( \bcancel{\rm{1} } \) 2 2 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1" that is understood to occupy the last "blank" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Exercise \(\PageIndex{2}\) Balance the following chemical equation by writing coefficients in the "blanks," as necessary. ___ \(\ce{NH_3} \left( g \right) +\) ___ \(\ce{O_2} \left( g \right) \rightarrow\) ___ \(\ce{N_2} \left( g \right) +\) ___ \(\ce{H_2O} \left( g \right)\) Answer In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of nitrogen, N, hydrogen, H, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced N 1 2 H 3 2 O 2 1 In order for a component of a reaction to be balanced, the reactants and products of the reaction must contain identical total quantities of that type of element or polyatomic ion. Therefore, none of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. None of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow or is represented in its atomic form. Therefore, the balancing process could be initiated by selecting any of these elements for further analysis. In order to balance nitrogen, N, a coefficient should be written in the "blank" that corresponds to nitrogen trihydride, NH3, on the left side of the equation, as fewer nitrogens are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{NH_3} \left( g \right) +\) ___ \(\ce{O_2} \left( g \right) \rightarrow\) ___ \(\ce{N_2} \left( g \right) +\) ___ \(\ce{H_2O} \left( g \right)\) As nitrogen trihydride, NH3, contains both nitrogen, N, and hydrogen, H, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements, which are obtained by multiplying the subscripts that are associated with each elemental symbol by the coefficient that was written, are reflected in the table that is shown below. Inserting this coefficient balances nitrogen, N, as intended. The quantities in which hydrogen, H, and oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced N \( \cancel{\rm{1} } \) 2 2 H \( \cancel{\rm{3} } \) 6 2 O 2 1 In order to balance hydrogen, H, a coefficient should be written in the "blank" that corresponds to water, H2O, on the right side of the equation, as fewer hydrogens are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 6, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{NH_3} \left( g \right) +\) ___ \(\ce{O_2} \left( g \right) \rightarrow\) ___ \(\ce{N_2} \left( g \right) +\) 3 \(\ce{H_2O} \left( g \right)\) As water, H2O, contains both hydrogen, H, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements, which are obtained by multiplying the subscripts that are associated with each elemental symbol by the coefficient that was written, are reflected in the table that is shown below. Inserting this coefficient balances hydrogen, H, as intended. The quantities in which oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced N \( \cancel{\rm{1} } \) 2 2 H \( \cancel{\rm{3} } \) 6 \( \bcancel{\rm{2} } \) 6 O 2 \( \bcancel{\rm{1} } \) 3 In order to balance oxygen, O, a coefficient should be written in the "blank" that corresponds to molecular oxygen, O2, on the left side of the equation, as fewer oxygens are present on this side of the reaction arrow. The value of this coefficient, 1.5, is determined by dividing the larger quantity of this element, 3, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{NH_3} \left( g \right) +\) 1.5 \(\ce{O_2} \left( g \right) \rightarrow\) ___ \(\ce{N_2} \left( g \right) +\) 3 \(\ce{H_2O} \left( g \right)\) As oxygen, O, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of nitrogen, N, or hydrogen, H, that are present on the reactant side of the equation. The updated quantity of this element, which was obtained by multiplying its subscript by the coefficient that was written, is reflected in the table that is shown below. Inserting this coefficient balances oxygen, O, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced N \( \cancel{\rm{1} } \) 2 2 H \( \cancel{\rm{3} } \) 6 \( \bcancel{\rm{2} } \) 6 O \( \bcancel{\rm{2} } \) 3 \( \bcancel{\rm{1} } \) 3 However, a fractional coefficient, 1.5, is written in the equation that is shown above. As a result, all of the coefficients in this equation, including the unwritten "1" that is understood to occupy the third blank, must be multiplied by 2, in order to cancel this half-fraction. The doubled coefficient values are reflected in the chemical equation that is shown below. 4 \(\ce{NH_3} \left( g \right) +\) 3 \(\ce{O_2} \left( g \right) \rightarrow\) 2 \(\ce{N_2} \left( g \right) +\) 6 \(\ce{H_2O} \left( g \right)\) By multiplying all of the coefficients in this equation by 2, the quantities in which nitrogen, N, hydrogen, H, and oxygen, O, are present in the equation have changed, as shown in the table below, but their relative ratios have not. Therefore, all of the components of this equation are still balanced. Element or Ion Reactants Products Balanced N \( \cancel{\rm{1} } \) \( \cancel{\rm{2} } \) 4 \( \cancel{\rm{2} } \) 4 H \( \cancel{\rm{3} } \) \( \cancel{\rm{6} } \) 12 \( \bcancel{\rm{2} } \) \( \cancel{\rm{6} } \) 12 O \( \bcancel{\rm{2} } \) \( \cancel{\rm{3} } \) 6 \( \bcancel{\rm{1} } \) \( \cancel{\rm{3} } \) 6 Finally, these coefficients cannot be divided, as they do not all share a common divisor that would result in the calculation of four whole number coefficients. Therefore, the final equation that is presented above is chemically-correct, as written. Exercise \(\PageIndex{3}\) Balance the following chemical equation by writing coefficients in the "blanks," as necessary. ___ \(\ce{C_1_2H_2_6} \left( l \right) +\) ___ \(\ce{O_2} \left( g \right) \rightarrow\) ___ \(\ce{CO_2} \left( g \right) +\) ___ \(\ce{H_2O} \left( g \right) + \ce{E}\) Answer This reaction is classified as a combustion because energy, "E," is generated as a product. Recall that coefficients are incorporated into a chemical equation in order to account for any relative differences between the chemical formulas of the reactants and products that are involved in the corresponding chemical reaction. However, energy is not a chemical material and, therefore, does not contain elements or polyatomic ions. As a result, there is no need to associate a balancing coefficient with the energy that is produced in a combustion reaction. In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of carbon, C, hydrogen, H, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Note that oxygen is present in both of the chemical products that are generated during this reaction. In order to determine the number of oxygens that are present on the product side of this equation, the subscripts on both oxygens, which are a "2" and an unwritten "1," respectively, must be added. Therefore, a total of three oxygens are initially present on the product side of this equation. Element or Ion Reactants Products Balanced C 12 1 H 26 2 O 2 3 In order for a component of a reaction to be balanced, the reactants and products of the reaction must contain identical total quantities of that type of element or polyatomic ion. Therefore, none of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. None of the unbalanced components in this reaction is represented in its atomic form. However, as stated above, oxygen is present in both of the chemical products that are generated during this reaction and, therefore, should not be the first element to be balanced in this equation. As neither carbon, C, nor hydrogen, H, is present in multiple formulas on the same side of the reaction arrow or is represented in its atomic form, the balancing process could be initiated by selecting either of these elements for further analysis. In order to balance carbon, C, a coefficient should be written in the "blank" that corresponds to carbon dioxide, CO2, on the right side of the equation, as fewer carbons are present on this side of the reaction arrow. The value of this coefficient, 12, is determined by dividing the larger quantity of this element, 12, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_1_2H_2_6} \left( l \right) +\) ___ \(\ce{O_2} \left( g \right) \rightarrow\) 12 \(\ce{CO_2} \left( g \right) +\) ___ \(\ce{H_2O} \left( g \right) + \ce{E}\) As carbon dioxide, CO2, contains both carbon, C, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements, which are obtained by multiplying the subscripts that are associated with each elemental symbol by the coefficient that was written, are reflected in the table that is shown below. As stated above, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances carbon, C, as intended. The quantities in which hydrogen, H, and oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 12 \( \cancel{\rm{1} } \) 12 H 26 2 O 2 \( \cancel{\rm{3} } \) 25 In order to balance hydrogen, H, a coefficient should be written in the "blank" that corresponds to water, H2O, on the right side of the equation, as fewer hydrogens are present on this side of the reaction arrow. The value of this coefficient, 13, is determined by dividing the larger quantity of this element, 26, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_1_2H_2_6} \left( l \right) +\) ___ \(\ce{O_2} \left( g \right) \rightarrow\) 12 \(\ce{CO_2} \left( g \right) +\) 13 \(\ce{H_2O} \left( g \right) + \ce{E}\) As water, H2O, contains both hydrogen, H, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements, which are obtained by multiplying the subscripts that are associated with each elemental symbol by the coefficient that was written, are reflected in the table that is shown below. Again, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances hydrogen, H, as intended. The quantities in which oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 12 \( \cancel{\rm{1} } \) 12 H 26 \( \bcancel{\rm{2} } \) 26 O 2 \( \cancel{\rm{3} } \) \( \bcancel{\rm{25} } \) 37 In order to balance oxygen, O, a coefficient should be written in the "blank" that corresponds to molecular oxygen, O2, on the left side of the equation, as fewer oxygens are present on this side of the reaction arrow. The value of this coefficient, 18.5, is determined by dividing the larger quantity of this element, 37, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_1_2H_2_6} \left( l \right) +\) 18.5 \(\ce{O_2} \left( g \right) \rightarrow\) 12 \(\ce{CO_2} \left( g \right) +\) 13 \(\ce{H_2O} \left( g \right) + \ce{E}\) As oxygen, O, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of carbon, C, or hydrogen, H, that are present on the reactant side of the equation. The updated quantity of this element, which was obtained by multiplying its subscript by the coefficient that was written, is reflected in the table that is shown below. Inserting this coefficient balances oxygen, O, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced C 12 \( \cancel{\rm{1} } \) 12 H 26 \( \bcancel{\rm{2} } \) 26 O \( \bcancel{\rm{2} } \) 37 \( \cancel{\rm{3} } \) \( \bcancel{\rm{25} } \) 37 However, a fractional coefficient, 18.5, is written in the equation that is shown above. As a result, all of the coefficients in this equation, including the unwritten "1" that is understood to occupy the first blank, must be multiplied by 2, in order to cancel this half-fraction. The doubled coefficient values are reflected in the chemical equation that is shown below. 2 \(\ce{C_1_2H_2_6} \left( l \right) +\) 37 \(\ce{O_2} \left( g \right) \rightarrow\) 24 \(\ce{CO_2} \left( g \right) +\) 26 \(\ce{H_2O} \left( g \right) + \ce{E}\) By multiplying all of the coefficients in this equation by 2, the quantities in which carbon, C, hydrogen, H, and oxygen, O, are present in the equation have changed, as shown in the table below, but their relative ratios have not. Therefore, all of the components of this equation are still balanced. Element or Ion Reactants Products Balanced C \( \cancel{\rm{12} } \) 24 \( \cancel{\rm{1} } \) \( \cancel{\rm{12} } \) 24 H \( \cancel{\rm{26} } \) 52 \( \bcancel{\rm{2} } \) \( \cancel{\rm{26} } \) 52 O \( \bcancel{\rm{2} } \) \( \cancel{\rm{37} } \) 74 \( \cancel{\rm{3} } \) \( \bcancel{\rm{25} } \) \( \cancel{\rm{37} } \) 74 Finally, these coefficients cannot be divided, as they do not all share a common divisor that would result in the calculation of four whole number coefficients. Therefore, the final equation that is presented above is chemically-correct, as written.
