Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.
ch₃(ch₂)₇ch₃(l) + o₂(g) → co₂(g) + h₂o(g)

Explanation:

Step1: Balance carbon atoms

The hydrocarbon \(CH_3(CH_2)_7CH_3\) (nonane, \(C_9H_{20}\)) has 9 carbon atoms. So, we put a coefficient of 9 in front of \(CO_2\).
\(C_9H_{20}(l)+O_2(g)\to9CO_2(g)+H_2O(g)\)

Step2: Balance hydrogen atoms

\(C_9H_{20}\) has 20 hydrogen atoms. So, we put a coefficient of 10 in front of \(H_2O\).
\(C_9H_{20}(l)+O_2(g)\to9CO_2(g)+10H_2O(g)\)

Step3: Balance oxygen atoms

On the right - hand side, the number of oxygen atoms in \(9CO_2\) is \(9\times2 = 18\), and in \(10H_2O\) is \(10\times1=10\). So, the total number of oxygen atoms on the right - hand side is \(18 + 10=28\). Since each \(O_2\) molecule has 2 oxygen atoms, we put a coefficient of 14 in front of \(O_2\).
\(C_9H_{20}(l)+14O_2(g)\to9CO_2(g)+10H_2O(g)\)

Answer:

\(1C_9H_{20}(l)+14O_2(g)\to9CO_2(g)+10H_2O(g)\)