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

2Fe+3S->Fe2S3

51.57 g.

Explanation:

The balanced equation: 2Fe + 3S → Fe₂S₃, it is clear that 2.0 moles of Fe react with 3.0 moles of S to produce 1.0 mole of Fe₂S₃. We need to calculate the no. of moles of Fe₂S₃:

n = mas/molar mass = (96.0 g) / (207.9 g/mol) = 0.46 mol.

Using cross multiplication we can get the no. of moles of Fe that isn eeded to produce 0.46 mol of Fe₂S₃:

2.0 moles of Fe produces → 1.0 mol of Fe₂S₃.

? moles of Fe produces → 0.46 mol of Fe₂S₃.

∴ The no. of moles of Fe needed = (2.0 mol) (0.46 mol) / (1.0 mol) = 0.92 mol.

Now, we can get the mass of 0.92 mol Fe:

mass = no. of moles x molar mass = (0.92 mol) (55.845 g/mol) = 51.57 g.