Question

(a) What volume of hydrogen sulphide at STP will burn in oxygen to yield 12.8 g sulphur dioxide according to the equation?
2H₂S + 3O₂ $\stackrel{}{\to }$ 2H₂O + 2SO₂
(H = 1, O = 16, S = 32)
(b) For the volume of hydrogen sulphide determined in (a) above, what volume of oxygen would be required for complete combustion?

Answer 1

(a) Let V be the volume of hydrogen sulphide required at STP.
Chemical equation for the reaction is:

2H₂S + 3O₂ $\to$ 2H₂O + 2SO₂

Number of moles in 12.8 g of SO₂ = $\frac{\mathrm{Mass}\mathrm{of}{\mathrm{SO}}_2}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{SO}}_2}=\frac{12.8}{64}=0.2$
According to the balanced chemical equation, two moles of SO₂ is produced by two moles of H₂S.
0.2 moles of SO₂ is produced by 0.2 moles of H₂S.
Number of moles of H₂S at STP = $\frac{\mathrm{Volume}\mathrm{of}{\mathrm{H}}_2\mathrm{S}}{\mathrm{Molar}\mathrm{Volume}\mathrm{of}{\mathrm{H}}_2\mathrm{S}}=\frac{V}{22.4}$
or
$$\begin{gathered} \frac{V}{22.4}=0.2 \\ V=4.48\mathrm{L} \end{gathered}$$

Thus, 4.48 litres of H₂S is required to produce 12.8 grams of SO₂.

(b) According to the balanced chemical equation, two moles of H₂S require three moles of O₂ for complete combustion.
In terms of volume of reacting gases, it can be said that:
Volume of O₂ required by 44.8 litres of H₂S = (3 $\times$ 22.4) L = 67.2 L
Volume of O₂ required by 4.48 litres of H₂S = $\frac{67.2\times 4.48}{44.8}=6.72\mathrm{L}$