Answer to Question #195119 in Chemistry for Dan

Question

Sulphuric acid (H2SO4) reacts with aluminium to produce aluminium sulphate

(Al2 (SO4)3 ) and hydrogen (gas). Write and balance the reaction. If 500 g of a 46% by

mass solution of sulphuric acid is combined with aluminium

a) How many moles of sulphuric acid will there be in 500 g of 46% solution of this

acid?

b) Calculate the moles of aluminium necessary to completely consume the moles of

sulphuric acid obtained in the previous section.

c) Determine the moles of aluminium sulphate that will be produced and express

this quantity in grams.

Answer 1

Solution:

Balanced chemical equation:

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

a)

Mass of H₂SO₄ = (Mass of H₂SO₄ solution × Mass percent) / 100%

Mass of H₂SO₄ = (500 g × 46%) / 100 % = 230 g

The molar mass of H₂SO₄ is 98.079 g/mol.

Hence,

(230 g H₂SO₄) × (1 mol H₂SO₄ / 98.079 g H₂SO₄) = 2.345 mol H₂SO₄

b)

(Moles of Al) / 2 = (Moles of H₂SO₄) / 3 (according to the equation above)

Moles of Al = 2 × Moles of H₂SO₄ / 3 = (2 × 2.345 mol) / 3 = 1.563 mol Al

c)

Moles of Al₂(SO₄)₃ = (Moles of H₂SO₄) / 3 (according to the equation above)

Moles of Al₂(SO₄)₃ = 2.345 mol / 3 = 0.782 mol Al₂(SO₄)₃

The molar mass of Al₂(SO₄)₃ is 342.15 g/mol.

Hence,

(0.782 mol Al₂(SO₄)₃) × (342.15 g Al₂(SO₄)₃ / 1 mol Al₂(SO₄)₃) = 267.56 g Al₂(SO₄)₃

Answers:

Balanced chemical equation: 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

a) 2.345 mol H₂SO₄

b) 1.563 mol Al

c) 0.782 mol Al₂(SO₄)₃ or 267.56 g Al₂(SO₄)₃