Question

How many millilitres of sulphur dioxide are formed when 12.5 g of iron sulphide
ore (pyrite) reacts with oxygen according to the equation at STP?​

Answer 1

Given info : 12.5 g of iron sulphide ore (Pyrite ) reacts with oxygen according to the equation at STP.

To find : volume of sulphur dioxide is ...

solution : mass of iron sulphide = 12.5 g

molar mass of iron sulphide = 120 g/mol

so no of moles of iron sulphide = 12.5/120 = 0.104167 mol

chemical equation of reaction of iron sulphide with oxygen is ..

4FeS₂ + 11O₂ ⇒2Fe₂O₃ + 8SO₂

here you see, 4 mol of FeS₂ gives 8 mole of sulphur dioxide.

⇒1 mol of FeS₂ = 8/4 mol = 2 mol of sulphur dioxide.

⇒0.104167 of FeS₂ = 2 × 0.104167 = 0.208334 mol of Sulphur dioxide.

at STP 1 mol = 22.4 L

so the mass of sulphur dioxide = 0.208334 × 22.4 L = 4.6666816 L = 4666.6816 ml

Therefore the volume of sulphur dioxide is 4666.6816 ml.

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