Question 6.7: Balance the following chemical equation. C2H6O + O2 → CO2 + ......

The element oxygen appears in four different places in this chemical equation. This means we do not want to start the balancing process with the element oxygen. Always start the balancing process with an element that appears only once on both the reactant and product sides of the equation.
Step 1:     Balancing of H atoms. There are six H atoms on the left and two H atoms on the right. Placing the coeffi cient 1 in front of C_{2}H_{6}O and the coeffi cient 3 in front of H_{2}O balances the H atoms at six on each side.
\quad\quad\quad\quad 1C_{2}H_{6}O + O_{2} \longrightarrow CO_{2} + {3}H_{2}O
Step 2:     Balancing of C atoms. An effect of balancing the H atoms at six (Step 1) is the setting of the C atoms on the left side at two. Placing the coeffi cient 2 in front of CO_{2} causes the carbon atoms to balance at two on each side of the chemical equation.
\quad\quad\quad\quad 1C_{2}H_{6}O + O_{2} \longrightarrow 2CO_{2} + {3}H_{2}O

Step 3:     Balancing of O atoms. The oxygen content of the right side of the chemical equation is set at seven atoms: four oxygen atoms from {2}CO_{2} and three oxygen atoms from {3}H_{2}O. To obtain seven oxygen atoms on the left side of the chemical equation, we place the coeffi cient 3 in front of O_{2}; 3O_{2} gives six oxygen atoms, and there is an additional O in 1C_{2}H_{6}O. The element oxygen is present in all four formulas in the chemical equation.
\quad\quad\quad\quad 1C_{2}H_{6}O + {3}O_{2} \longrightarrow {2}CO_{2} + {3}H_{2}O
Step 4:     Final check. The equation is balanced. There are two carbon atoms, six hydrogen atoms, and seven oxygen atoms on each side of the chemical equation.
\quad\quad\quad\quad C_{2}H_{6}O + {3}O_{2} \longrightarrow {2}CO_{2} + {3}H_{2}O