Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.

ch₃(ch₂)₆ch₃(g) + o₂(g) → co₂(g) + h₂o(g)

Explanation:

Step1: Balance carbon atoms

The hydrocarbon \(CH_3(CH_2)_6CH_3\) (which has \(C_8H_{18}\) as its molecular formula) has 8 carbon atoms. So, we put a coefficient of 8 in front of \(CO_2\):
\(C_8H_{18}(g)+O_2(g)\to8CO_2(g)+H_2O(g)\)

Step2: Balance hydrogen atoms

\(C_8H_{18}\) has 18 hydrogen atoms. So, we put a coefficient of 9 in front of \(H_2O\):
\(C_8H_{18}(g)+O_2(g)\to8CO_2(g)+9H_2O(g)\)

Step3: Balance oxygen atoms

On the right - hand side, the number of oxygen atoms is \(8\times2 + 9\times1=16 + 9=25\). So, we put a coefficient of \(\frac{25}{2}\) in front of \(O_2\):
\(C_8H_{18}(g)+\frac{25}{2}O_2(g)\to8CO_2(g)+9H_2O(g)\)
Since we need whole - number coefficients, we multiply each coefficient by 2:
\(2C_8H_{18}(g)+25O_2(g)\to16CO_2(g)+18H_2O(g)\)

Answer:

\(2CH_3(CH_2)_6CH_3(g)+25O_2(g)\to16CO_2(g)+18H_2O(g)\)