Question

The volume of oxygen required for complete combustion of 0.25 mole of methane at STP is ______.

पर्याय

• 22.4 dm³

• 11.2 dm³

• 7.46 dm³

• 5.6 dm³

Answer 1

The volume of oxygen required for complete combustion of 0.25 mole of methane at STP is 11.2 dm³.

Explanation:

Volume of methane = 0.25 mol = 5.6 L
[Multiply mole value by molar volume constant, 22.4 L]

Combustion of methane (CH₄)

\[\ce{\underset{Mathane}{CH4}+2O2->CO2+2H2O}\]

As, we know that, the volume of 1 mole of gas is 22.4 L.
Amount of oxygen is required for the combustion of 22.4 L of methane = (2 × 22.4) L = 44.8 L

Amount of oxygen required for the combustion of 5.6 L of methane

`=44.8/22.4xx5.6=11.2"L"=11.2"dm"^3`  [∵ 1L = 1dm³]

Hence, 11.2 dm³ of oxygen required for the complete combustion of 0.25 mole of methane.