Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:

2H₂S + 3O₂ ⟶ 2H₂O + 2SO₂

Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at S.T.P. which will complete the combustion of hydrogen sulphide determined in (litres).

$\begin{matrix} 2\text{H}$2\text{S} & + &3\text{O}2 & \longrightarrow & 2\text{H}2\text{O} & + & 2\text{SO}2 \ 2\text{ Vol.} && 3\text{ Vol.} && && 2\text{ Vol.} \ 2[2(1) + 32] && 3[2(16)] &&&& 2[32 + 2(16)] \ 68 \text{ g} && 96 \text{ g} &&&& 128 \text{ g} \end{matrix}

128 g of SO₂ has volume 2 × 22.4 litres

∴ 12.8 g of SO₂ has volume =

$\dfrac{2 × 22.4}{128}$ x 12.8 = 4.48 L

Volume of oxygen = ?

2 × 22.4 L H₂S requires = 3 × 22.4 litre of oxygen

∴ 4.48 L H₂S will require

$\dfrac{3 × 22.4}{2 × 22.4 }$ x 4.48 = 6.72 L of oxygen.