Answer to Question #295069 in Chemistry for ella

Question

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

2Fe + 3S ---->  Fe₂S₃

Answer 1

Solution:

Calculate the moles of iron(III) sulfide (Fe₂S₃):

The molar mass of Fe₂S₃ is 207.9 g/mol

Hence,

Moles of Fe₂S₃ = (96 g Fe₂S₃) × (1 mol Fe₂S₃ / 207.9 g Fe₂S₃) = 0.4618 mol Fe₂S₃

Balanced chemical equation:

2Fe + 3S → Fe₂S₃

According to stoichiometry:

2 mol of Fe produces 1 mol of Fe₂S₃

X mol of Fe produces 0.4618 mol of Fe₂S₃:

Thus:

Moles of Fe = X = (0.4618 mol Fe₂S₃) × (2 mol Fe / 1 mol Fe₂S₃) = 0.9236 mol Fe

Calculate the mass of iron (Fe):

The molar mass of Fe is 55.845 g/mol

Hence,

Mass of Fe = (0.9236 mol Fe) × (55.845 g Fe / 1 mol Fe) = 51.58 g Fe

Answer: 51.58 grams of iron (Fe) are needed.