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Question

What volume of Sulphur dioxide (at s.t.p) would be liberated by roasting 30 g of pyrites?


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Solution

Step 1:Write the chemical reaction

4FeS2 + 11O2 2Fe2O3 + 8SO2

Iron disulphide oxygen Iron(III) oxide Sulphur dioxide

Given:

(S= 32, Fe= 56, molar volume of gas is 22.4 litres at s.t.p)

Step 2: Find the molar mass

Molar mass of FeS2=56 + 2 x 32

=56 + 64

= 120 g/mol

According to the equation, the molecular weight of 4 moles of FeS2= 4 x 120

= 480 g/mol

The molar volume of gas is 22.4 litres at s.t.p

So, the molar volume of 8 SO2= 8 x 22.4 litres

= 179.2 litres

Step 3 Find the volume:

480 g of FeS2 on roasting gives 179.2 litres of SO2 at S.T.P.

Therefore, 30 g of FeS2 on roasting will give = 179.2 x 30480 = 11.2 litres of SO2 at S.T.P.

Step 4: Result

Hence, 11.2 litres of the volume of Sulphur dioxide (at s.t.p) would be liberated.


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