Question

A sample weighing 3.102 g is a mixture of Fe2O3 (molar mass = 159.69 g/mol) and Al2O3 (molar mass = 101.96 g/mol). When heat and a stream of H2 gas is applied to the sample, the Fe2O3 reacts to form metallic Fe and H2O(g). The Al2O3 does not react. If the sample residue (the solid species remaining after the reaction) weighs 2.413 g, what is the mass fraction of Fe2O3 in the original sample?

Answer 1

Answer:

The mass fraction of Fe2O3 in the original sample is 74.13 %

Explanation:

The reaction is given by:

Fe₂O₃ (s) + 3 H₂ (g) ⇒ 2 Fe (s) + 3 H₂O (g)

The initial state fo the mixture is:

Mass Fe₂O₃ + Mass Al₂O₃ = 3.102 g

Moles of Fe₂O₃ × Molar Mass Fe₂O₃ + Moles of Al₂O₃ × Molar Mass Al₂O₃= 3.102 g

Moles Fe₂O₃ × MM Fe₂O₃ + Moles Al₂O₃ × MM Al₂O₃= 3.102 g

The final state of the mixture is:

Mass Fe + Mass Al₂O₃ = 2.413 g

Moles of Fe × Molar Mass Fe + Moles of Al₂O₃ × Molar Mass Al₂O₃= 2.413 g

Since 2 moles of Fe are generated for each mole of Fe2O3, the equation can be written this way:

2 × (Moles of Fe₂O₃) × Molar Mass Fe + Moles of Al₂O₃ × Molar Mass Al₂O₃= 2.413 g

2×Moles Fe₂O₃× MM Fe + Moles Al₂O₃ × MM Al₂O₃= 2.413 g

Now we perform a subtraction between the two equations:

Moles Fe₂O₃ × MM Fe₂O₃ + Moles Al₂O₃ × MM Al₂O₃ -2×Moles Fe₂O₃× MM Fe - Moles Al₂O₃ × MM Al₂O₃= 3.102 g - 2.413 g

Moles Fe₂O₃ × MM Fe₂O₃ -2×Moles Fe₂O₃× MM Fe = 0.689 g

Moles Fe₂O₃ ×(MM Fe₂O₃ -2× MM Fe)= 0.689 g

Moles Fe₂O₃ ×(159.69 g/mol - 2× 55.85 g/mol)= 0.689 g

Moles Fe₂O₃ (47.99 g/mol)= 0.689 g

Moles Fe₂O₃ = 0.689 g ÷ (47.99 g/mol)

Moles Fe₂O₃ = 0.0144 mol

The mass fraction of Fe₂O₃ is given by:

% Fe₂O₃= $\frac{0.0144 mol*159.69g/mol}{3.102g}*100=74.13%$