Question

Examine the statement and balanced equation.

Pure elemental iron undergoes a single replacement reaction with sulfuric acid (H2SO4) to produce iron(III) sulfate (Fe2(SO4)3) and hydrogen gas.

2Fe + 3H2SO4 → Fe2(SO4)3 + 3H2
(M Fe = 55.8 g/mol, M H2SO4 = 98.1 g/mol, M Fe2(SO4)3 = 399.9 g/mol, M H2 = 2.0 g/mol)

If 3.21 x 1012 mol of iron(III) sulfate are formed, how many g of it will be produced?

Answer 1

The answer to your question is: 1.28 x 10 ¹⁵ g of Fe₂(SO₄)₃

Step-by-step explanation:

Single replacement reaction:

2Fe + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂

M Fe = 55.8 g/mol

M H₂SO₄ = 98.1 g/mol

M Fe₂(SO₄)₃ = 399.9 g/mol

M H₂ = 2.0 g/mol

Fe₂(SO₄)₃ = 3.21 x 10¹² mol

Fe₂(SO₄)₃ = ? g

1 mol of Fe₂(SO₄)₃ ----------------- 399.9 g/mol of Fe₂(SO₄)₃

3.21 x 10¹² mol Fe₂(SO₄)₃ -------- x

x = (3.21 x 10¹² x 399.9 ) / 1

x = 1.28 x 10 ¹⁵ g of Fe₂(SO₄)₃