Question

Consider the following reaction:
$8Fe+S_8\to 8FeS$
If we begin with 293 g of Fe and 17.2 g of $S_8$, how many grams of FeS will be produced?

• A. 47.17 g
• B. 52.03 g
• C. 42.30 g
• D. 45.00 g

Answer 1

The correct option is A 47.17 g

Moles of Fe = $\frac{293}{56}=5.23$ moles
Moles of $S_8=\frac{17.2}{256}=0.067$ moles
$8Fe+S_8\to 8FeS$
8 moles of Fe react with 1 mole of $S_8$
Therefore, 5.23 moles of Fe will react with $\frac{1}{8}\times 5.23=0.65$ moles of $S_8$
So, $S_8$ is the limiting reagent.
1 mole of $S_8$ produces 8 moles of FeS
0.067 moles will produce = $8\times 0.067=0.536$ moles
Mass of FeS $=0.536\times$ molar mass of FeS
Molar mass of FeS is 88
Mass of FeS = $0.536\times 88$
Mass of FeS = 47.17 g