Question 6.2: Balancing a Chemical Equation The major ingredient in ordina......
Balancing a Chemical Equation
The major ingredient in ordinary safety matches is potassium chlorate, KClO_3, a substance that can act as a source of oxygen in combustion reactions. Its reaction with ordinary table sugar (sucrose, C_{12}H_{22}O_{11}), for instance, occurs violently to yield potassium chloride, carbon dioxide, and water. Write a balanced equation for the reaction.
Step 1 Write the unbalanced equation, making sure the formulas for all substances are correct:
\mathrm{KClO}_3+\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow \mathrm{KCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \text { Unbalanced }Step 2 Find coefficients to balance the equation by starting with the most complex substance (sucrose) and considering one element at a time. Since there are 12 C atoms on the left and only 1 on the right, we can balance for carbon by adding a coefficient of 12 to CO_2 on the right:
\mathrm{KClO}_3+\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow \mathrm{KCl}+12 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \text { Balanced for } \mathrm{C}Since there are 22 H atoms on the left and only 2 on the right, we can balance for hydrogen by adding a coefficient of 11 to H_2O on the right:
\mathrm{KClO}_3+\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow \mathrm{KCl}+12 \mathrm{CO}_2+11 \mathrm{H}_2 \mathrm{O} \text { Balanced for } \mathrm{C} \text { and } \mathrm{H}There are now 35 O atoms on the right but only 14 on the left (11 in sucrose and 3 in KClO_3). Thus, 21 oxygens must be added on the left. We can do this without disturbing the C and H balance by adding 7 more KClO_3 , giving a coefficient of 8 for KClO_3 on the left:
8 \mathrm{KClO}_3+\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow \mathrm{KCl}+12 \mathrm{CO}_2+11 \mathrm{H}_2 \mathrm{O} \text { Balanced for } \mathrm{C}, \mathrm{H} \text {, and } \mathrm{O}Potassium and chlorine can both be balanced by adding a coefficient of 8 to KCl on the right:
8 \mathrm{KClO}_3+\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow 8 \mathrm{KCl}+12 \mathrm{CO}_2+11 \mathrm{H}_2 \mathrm{O} \text { Balanced for } \mathrm{C}, \mathrm{H}, \mathrm{O}, \mathrm{K} \text {, and } \mathrm{Cl}Steps 3 and 4 The coefficients in the balanced equation are already reduced to their smallest whole-number values, and a check shows that the numbers and kinds of atoms are the same on both sides of the equation.
