Question

Find volume of oxygen required to burn completely 112 litres of methane at STP?

Answer 1

ch4 molecular weight = 16g

16g- 11.2lit_22.4 li=?

16×22.4÷11.2 =32

Answer 2

The volume of oxygen gas required is 224 L

Explanation:

We are given:

Volume of methane = 112 L

At STP:

1 mole of a gas occupies 22.4 L of volume

The chemical equation for the combustion of methane follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

22.4 L of methane reacts with (2 × 22.4) L of oxygen gas

So, 112 L of methane will react with = $\frac{(2\times 22.4)}{22.4}\times 112=224L$ of oxygen gas

Learn more about stoichiometry and STP:

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