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Question

Calculate the number of moles of Fe2O3 when 600 g of FeS2, reacts with 800 g of O2.
4FeS2+11O22Fe2O3+8SO2
(Given: Molar mass of Fe is 56 g/mol)

A
1 mol
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B
1.5 mol
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C
2 mol
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D
2.5 mol
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Solution

The correct option is D 2.5 mol
Given balanced chemical reaction is,
4FeS2+11O22Fe2O3+8SO2
We know,
Molar mass of FeS2=(56+32×2) g/mol=120 g/mol

Molar mass of O2 = 32 g/mol
600 g of FeS2=600120 mol FeS2=5 mol of FeS2

800 g O2=80032 mol of O2=25 mol of O2

From the balanced reaction,
11 moles of O2 reacts with 4 moles of FeS2
Therefore, 25 moles of O2 will react with
=411×25 mol FeS2=9.09 mol FeS2

So, FeS2 is the limiting reagent here.

Therefore, moles of Fe2O3 produced
=24×5 mol = 2.5 mol.

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