A 2.00 L bottle contains oxygen gas at STP. It is used to completely combust methane according to the reaction ${\text{CH}}_4+2{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$. Assuming complete consumption of the oxygen, calculate (a) the volume of methane required, (b) the volume of carbon dioxide produced, both at STP. (Take $1\text{mol}$ gas = $22.4\text{L}$ at STP.)

Step-by-step solution

1. Convert the given volume of ${\text{O}}_2$ to moles: $\frac{2.00\text{L}}{22.4{\text{L mol}}^{-1}}=0.0893\text{mol}$.
2. From the balanced equation, 2 mol ${\text{O}}_2$ react with 1 mol ${\text{CH}}_4$; therefore, moles of ${\text{CH}}_4$ needed = $\frac{0.0893}{2}=0.0446\text{mol}$.
3. Convert moles of ${\text{CH}}_4$ back to volume at STP: $0.0446\text{mol}\times 22.4{\text{L mol}}^{-1}=1.00\text{L}$ (rounded to three significant figures).
4. The same mole amount of ${\text{CO}}_2$ is produced (1 mol ${\text{CO}}_2$ per 1 mol ${\text{CH}}_4$).
5. Convert moles of ${\text{CO}}_2$ to volume: $0.0446\text{mol}\times 22.4{\text{L mol}}^{-1}=1.00\text{L}$.
6. Thus, (a) $1.0\text{L}$ of methane is required and (b) $1.0\text{L}$ of carbon dioxide is formed.

Concept tested: Gas volume calculations using stoichiometry

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