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

The correct option is A 16
The balanced chemical equation for the combustion of methane is given below:
${CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (g)$

From the above reaction,

1 mole of ${CH}_{4}$ requires 2 moles of $O_{2}$ to complete the reaction.

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

22.4 L of ${CH}_{4}$ requires $2 \times$ 22.4 L = 44.8 L of $O_{2}$ .

So, 1 L of ${CH}_{4}$ requires $\frac{44.8}{22.4}$ = 2 L of $O_{2}$ .

Hence, 8 L of ${CH}_{4}$ requires $2 \times$ 8 L = 16 L of $O_{2}$ .
16 L of $O_{2}$ is required to complete the combustion of 8 L of methane.

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Calculate volume of oxygen at STP required to complete the combustion of 8 L of methane.16

Calculate volume of oxygen at STP required to complete the combustion of 8 L of methane.
16