Question

Calculate the volume of oxygen required for the complete combustion of $0.25{dm}^3$ of methane at STP.

Answer 1

Given volume of methane $=0.25{dm}^3=0.25L$

Combustion of methane $\left(C{H}_4\right)$-

$\underset{\left(\text{methane}\right)}{C{H}_4}+2O_2⟶C{O}_2+2H_{2}O$

As we know that volume of 1 mole of gas is 22.4 L.

Amount of oxygen required for the combustion of 22.4 L of methane $=(2\times 22.4)L=44.8L$

$\therefore$ Amount of oxygen required for the combustion of 0.25 L of methane $=\frac{44.8}{22.4}\times 0.25=0.5L$

Hence, 0.5 L of oxygen required for the complete combustion of $0.25{dm}^3$ of methane.