-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 -8594- -160- -160- -160- -160- -160- -160- -160- -160- -160- -160- -160- -160- -160- 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-

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Mole Concept in Balanced Reaction

(a) What volu...

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 $\overset{}{\to}$ 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?

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Solution

(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 = $\frac{Mass of {SO}_{2}}{Molecular mass of {SO}_{2}} = \frac{12.8}{64} = 0.2$
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 = $\frac{Volume of H_{2} S}{Molar Volume of H_{2} S} = \frac{V}{22.4}$
or
$\frac{V}{22.4} = 0.2 V = 4.48 L$

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 = $\frac{67.2 \times 4.48}{44.8} = 6.72 L$

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