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Question

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:
2H2S + 3O2 2H2O + 2SO2
(i). calculate the volume of hydrogen suulphide at STP.
Also, (ii). calculate the volume of oxygen required at STP. Which will complete the combustion of hydrogen sulphide determined in litres.

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Solution

The molar mass of SO2 is 64 g/mol, it means 1 mol of SO2 has the mass 64 grams.

At STP condition, 1 mol of gas is equal to the volume of 22.4 L. In another way, 64 g of the SO2 is equal to 22.4 L.

Let us compare the balanced equation.

2H2S + 3O2 2H2O + 2SO2

Find the volume of SO2

12.8 g of SO2×( 22.4 L of SO264g SO2) = 4.48 L O2 gas

(i). Find the volume of H2S

4.48 L of SO2 will require = 4.48 L of SO2×( 2 L of H2S2 L SO2) = 4.48 L O2 gas

(ii). Find the volume of O2Note" Please use 4.48 L of H2S instead of 0.48 L of H2S .

4.48 L of SO2 will require = 4.48 L of SO2×( 3 L of O22 L SO2) = 6.72 L O2 gas

(iii).It is 6.72 L of O2 to burn completely the 4.48 L of H2S.


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