Question

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

$$2\mathrm{H}{\phantom{A}}_2\mathrm{S}+3\mathrm{O}{\phantom{A}}_2⟶2\mathrm{H}{\phantom{A}}_2\mathrm{O}+2\mathrm{SO}{\phantom{A}}_2$$

Calculate the volume of hydrogen sulphide at STP. Also, calculate the volume of oxygen required at STP which will complete the combustion of hydrogen sulphide determined in litres.

Answer 1

$$2\mathrm{H}{\phantom{A}}_2\mathrm{S}+3\mathrm{O}{\phantom{A}}_2⟶2\mathrm{H}{\phantom{A}}_2\mathrm{O}+2\mathrm{SO}{\phantom{A}}_2$$

At S.T.P.

(i) 2 vol. of H₂S = 2 × 22.4 = 44.8 litres

2SO₂ = 2[32 + (2 × 16)] = 128 g

128 g of SO₂ needs H₂S = 44.8 litres

∴ 12.8 g of SO₂ needs H₂S = ${\displaystyle \frac{44.8\times 12.8}{128}}$

= 4.48 litres

(ii) Also, 2 vols. of H₂S require O₂ = 3 volume

∴ 4.48 litres of H₂S requires O₂ = ${\displaystyle \frac{3}{2}\times 4.48}$

= 6.72 litres