Question

Calculate volume of oxygen at STP required to complete the combustion of 8 L of methane.

• A. 16

Answer 1

The correct option is A 16
The balanced chemical equation for the combustion of methane is given below:
${\mathrm{CH}}_4\left(g\right)+2O_2\left(g\right)\to {\mathrm{CO}}_2\left(g\right)+2H_{2}O\left(g\right)$

From the above reaction,

1 mole of ${\text{CH}}_4$ requires 2 moles of ${\text{O}}_2$ to complete the reaction.

Since, 1 mole of any gas at STP occupies 22.4 L, we can say

22.4 L of ${\text{CH}}_4$ requires $2\times$22.4 L = 44.8 L of ${\text{O}}_2$.

So, 1 L of ${\text{CH}}_4$ requires $\frac{44.8}{22.4}$ = 2 L of ${\text{O}}_2$.

Hence, 8 L of ${\text{CH}}_4$ requires $2\times$ 8 L = 16 L of ${\text{O}}_2$ .
16 L of ${\text{O}}_2$ is required to complete the combustion of 8 L of methane.