1.Let’s find the amount of iron (II) sulfide substance.
M (FeS) = 56 + 32 = 88 g / mol.
n = m: M.
n = 44 g / mol: 88 g / mol = 0.5 mol.
Fe + S = FeS.
2.According to the reaction equation, 1 mole of iron accounts for 1 mole of iron sulfide. Substances are in quantitative ratios of 1: 1. This means that their amount of substances will be the same.
n (Fe) = 0.5 mol.
3.Let’s find the mass of iron by the formula:
m = nM,
M (Fe) = 56 g / mol.
m = 56 g / mol х0.5 mol = 28 g.
Answer: 28 g.
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