Methane burns in oxygen according to the balanced equation CH_4 + 2O_2 → CO_2 + 2H_2O. Calculate the volume of oxygen at STP required to completely combust 16 g of methane.

Methane burns in oxygen according to the balanced equation $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ . Calculate the volume of oxygen at STP required to completely combust $16 g$ of methane.

medium 2 marks Class 10 · Science · Carbon and its Compounds

Answer

44.8 L

The correct answer is 44.8 L — Thus, $44.8 L$ of oxygen is needed.

Find the number of moles of methane: $n_{C H_{4}} = \frac{16 g}{M _{C H_{4}}} = \frac{16}{16} = 1 mol$ (Molar mass of CH $_{4}$ = 16 g·mol $^{- 1}$ ).
From the balanced equation, 1 mol of CH $_{4}$ needs 2 mol of O $_{2}$ . Hence $n_{O_{2}} = 2 \times 1 = 2 mol$ .
At STP, 1 mol of any gas occupies $22.4 L$ . Therefore, volume of O $_{2}$ required $V = n_{O_{2}} \times 22.4 = 2 \times 22.4 = 44.8 L$ .
Thus, $44.8 L$ of oxygen is needed.

Common mistake: Using the molar volume of

24 L

(room temperature) instead of

22.4 L

for STP, which gives an incorrect answer.

Concept tested: Stoichiometry of combustion reactions and use of molar volume at STP.