High School

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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.

\[ 2\text{Fe} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2 \]

(Molar masses: Mₓ(Fe) = 55.8 g/mol, Mₓ(H₂SO₄) = 98.1 g/mol, Mₓ(Fe₂(SO₄)₃) = 399.9 g/mol, Mₓ(H₂) = 2.0 g/mol)

If [tex]3.21 \times 10^{12}[/tex] mol of iron(III) sulfate are formed, how many grams of it will be produced?

Answer :

Answer:

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

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

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