What mass of sulfur reacted with 28 g of iron if 44 g of the FeS compound were formed?

Data: mFe (weight of iron taken) = 28 g; mFeS is the mass of the obtained iron sulfide (mFeS = 44 g).

Const: MFe – molar mass of iron (MFe = 55.845 g / mol ≈ 56 g / mol); MFeS – molar mass of iron sulfide (MFeS = 87.910 g / mol ≈ 88 g / mol).

To find out the required mass of the reacted sulfur, we apply the formula (mass conservation): mFeS = mFe + mS, from where we express: mS = mFeS – mFe = 44 – 28 = 16 g.

Checking.

1) Ur-tion of the reaction: Fe (iron) + S (sulfur) = FeS (iron sulfide).

2) Amount of iron sulfide substance νFeS = mFeS / MFeS = 44/88 = 0.5 mol

3) Amount of iron substance: νFe = mFe / MFe = 28/56 = 0.5 mol (all iron reacted, right).

Answer: Sulfur weighing 16 g should have reacted with 28 iron.