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Question

How many moles of Zn[FeS2]2 can be made from 2 g of Zn, 3 g of Fe and 4 g of S?
Zn+2Fe+4SZn[FeS2]2
(Given: Molar mass of Fe = 56 g/mol and molar mass of Zn = 65 g/mol)

A
0.0157 mol
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B
0.0265 mol
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C
0.2650 mol
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D
0.0538 mol
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Solution

The correct option is B 0.0265 mol
The given reaction is,
Zn+2Fe+4SZn[FeS2]2
So, from the balanced reaction,
1 mol of Zn reacts with 2 mol of Fe and 4 mol of S to give 1 mol of Zn[FeS2]2
Finding moles of reactants
2 g of Zn=265 mol of Zn = 0.030 mol of Zn
3 g of Fe=356 mol of Fe = 0.053 mol of Fe
4 g of S = 4/32 moles of S = 0.125 mol of S
We know, 1 mol of Zn reacts with 2 mol of Fe
So, 0.03 mol of Zn will react with 0.06 mol of Fe.
Here we have only 0.053 mol of Fe.
Again, 2 mol of Fe react with 4 mol of S.
So, 0.053 mol of Fe will react with 0.106 mol of S.
Thus, here Fe is the limiting reagent.
From the balanced chemical reaction,
2 mol of Fe react to give 1 mol of Zn[FeS2]2
Thus, 0.053 mol of Fe react to give 0.0265 mol of Zn[FeS2]2.

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