How many moles of $Zn[Fe{S}_2{]}_2$ can be made from 2 g of $Zn$, 3 g of $Fe$ and 4 g of $S$?
$Zn+2Fe+4S\to Zn[Fe{S}_2{]}_2$
(Given: Molar mass of $Fe$ = 56 g/mol and molar mass of $Zn$ = 65 g/mol)

The correct option is B 0.0265 mol
The given reaction is,
$Zn+2Fe+4S\to Zn[Fe{S}_2{]}_2$
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 $Zn[Fe{S}_2{]}_2$
Finding moles of reactants
2 g of $Zn=\frac{2}{65}$ mol of Zn = 0.030 mol of Zn
3 g of $Fe=\frac{3}{56}$ 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 $Zn[Fe{S}_2{]}_2$
Thus, 0.053 mol of Fe react to give 0.0265 mol of $Zn[Fe{S}_2{]}_2$.