Answer:
51.69 g of Fe
Step-by-step explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2Fe + 3S —> Fe₂S₃
Next, we shall determine the mass of Fe that reacted and the mass of Fe₂S₃ produced from the balanced equation. This can be obtained as follow:
Molar mass of Fe = 56 g/mol
Mass of Fe from the balanced equation = 2 × 56 = 112 g
Molar mass of Fe₂S₃ = (2×56) + (3×32)
= 112 + 96
= 208 g/mol
Mass of Fe₂S₃ from the balanced equation = 1 × 208 = 208 g
SUMMARY:
From the balanced equation above,
112 g of Fe reacted to produce 208 g of Fe₂S₃.
Finally, we shall determine the mass of Fe needed to produce 96 g of Fe₂S₃. This can be obtained as follow:
From the balanced equation above,
112 g of Fe reacted to produce 208 g of Fe₂S₃.
Therefore, Xg of Fe will react to produce 96 g of Fe₂S₃ i.e
Xg of Fe = (112 × 96)/208
Xg of Fe = 51.69 g
Thus, 51.69 g of Fe is needed for the reaction.