Question

Propane ${\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)$, is used in gas grills to provide heat for cooking. What volume of ${\mathrm{O}}_2$(g) measured at $25°\mathrm{C}$ and $760\mathrm{torr}$ is required to react with $2.7\mathrm{L}$ of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

Answer 1

Step 1: Balanced chemical equation

The balanced chemical reaction for the combustion of propane:

$\underset{\mathrm{Propane}}{{\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}}{5{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{2{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$

According to this balanced chemical equation, One mole ${\mathrm{C}}_3{\mathrm{H}}_8$ will react with five moles of${\mathrm{O}}_2$.

Step 2: Combustion of propane

Here, One volume of ${\mathrm{C}}_3{\mathrm{H}}_8$ will react with five volumes of ${\mathrm{O}}_2$.

So, the volume of ${\mathrm{O}}_2$ required to react with $2.7\mathrm{L}$ propane is,

$2.7\mathrm{L}{\mathrm{C}}_3{\mathrm{H}}_8\times \frac{5\mathrm{L}{\mathrm{O}}_2}{1\mathrm{L}{\mathrm{C}}_3{\mathrm{H}}_8}=13.5\mathrm{L}{\mathrm{O}}_2$

Therefore, the volume of ${\mathrm{O}}_2$ required to react with $2.7\mathrm{L}$ of propane is $13.5\mathrm{L}{\mathrm{O}}_2$.