Answer to Question #257057 in Chemistry for Marie

Question

the reaction of finely divided aluminum and iron(III) oxide, Fe2O3, is called the thermite reaction. it produces a tremendous amount of heat, making the welding od railroad track possible. the reaction of 500.0 grams of aluminum and 500.0 grams of iron(iii) oxide produces 166.5 grams iron. calculate the mass of iron that should be released by this reaction.

Answer 1

Solution:

The molar mass of Al is 26.98 g/mol

The molar mass of Fe is 55.845 g/mol

The molar mass of Fe₂O₃ is 159.69 g/mol

Calculate moles of each reactant:

(500.0 g Al) × (1 mol Al / 26.98 g Al) = 18.53 mol Al

(500.0 g Fe₂O₃) × (1 mol Fe₂O₃ / 159.69 g Fe₂O₃) = 3.13 mol Fe₂O₃

Balanced chemical equation:

2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s)

According to the chemical equation above:

2 mol of Al reacts with 1 mol of Fe₂O₃

Thus, 18.53 mol of Al reacts with:

(18.53 mol Al) × (1 mol Fe₂O₃ / 2 mol Al) = 9.265 mol Fe₂O₃

However, initially there is 3.13 mol of Fe₂O₃ (according to the task).

Therefore, Fe₂O₃ acts as limiting reagent and Al is excess reagent.

According to stoichiometry:

1 mol of Fe₂O₃ produces 2 mol of Fe

Thus, 3.13 mol of Fe₂O₃ produces:

(3.13 mol Fe₂O₃) × (2 mol Fe / 1 mol Fe₂O₃) = 6.26 mol Fe

Calculate the mass of Fe:

(6.26 mol Fe) × (55.845 g Fe / 1 mol Fe) = 349.59 g Fe = 349.6 g Fe

The mass of iron (Fe) is 349.6 grams

Answer: 349.6 grams of iron (Fe) should be released by this reaction.