What mass of iron is needed to react with sulfur in order to produce 96 grams of iron (III) sulfide according to the following equation?

51.69 g of Fe

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.

fe 55.85 g/mol and h2o 18.02 g/mol.