Question

(c) Calculate the volume of oxygen required for the complete combustion of $8.8\text{g}$ propane$\left({\mathrm{C}}_3{\mathrm{H}}_8\right)$.

[Atomic mass: $\mathrm{C}=12$, $\mathrm{O}=16$, $\mathrm{H}=1$, Molar volume$=22.4{\text{dm}}^3$ at STP]

Answer 1

Given information

$8.8\text{g}$ of propane$\left({\mathrm{C}}_3{\mathrm{H}}_8\right)$

S.T.P

• S.T.P indicates standard temperature and pressure
• At S.T.P, one mole of a gas has a volume of $22.4\text{L}$ or $22.4{\text{dm}}^3$.
• The mass of $22.4\text{L}$ of a gas is equal to its gram molecular mass.

Balanced chemical equation

• Propane $\left({\mathrm{C}}_3{\mathrm{H}}_8\right)$ reacts with oxygen $\left({\mathrm{O}}_2\right)$ and results in the formation of carbon dioxide $\left({\mathrm{CO}}_2\right)$ and water $\left({\mathrm{H}}_2\mathrm{O}\right)$.
• The balanced chemical reaction is as follows:

${\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2\to 3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}$

Calculation of molecular weight of propane

The molecular weight of propane

$$\begin{gathered} =3\mathrm{C}+8\mathrm{H} \\ =\left(3\times 12+8\times 1\right)\mathrm{g} \\ =44\mathrm{g} \end{gathered}$$

Calculation of required volume of oxygen

From the balanced equation, it can be said that $44\text{g}$ of propane requires $5\times 22.4\text{L}$ of oxygen at STP.

Therefore, $8.8\text{g}$ of propane will require $\frac{5\times 22.4\text{×8.8}}{44}=22.4\text{L}$ of oxygen.

Hence, $22.4\text{L}$ volume of oxygen is required for the complete combustion of $8.8\text{g}$ of propane.