Question

Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2

Answer 1

Answer:

$\large \boxed{\text{a)188.4 g; b) 98.67 \, \% }}$

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:                      98.08           392.18

             2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

• Use the molar mass of H₂SO₄ to convert  the mass of H₂SO₄ to moles of H₂SO₄
• Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃
• Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

$\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}$

(ii) Moles of H₂SO₄

$\text{Moles of H _{2} SO}_{4} = \text{141.36 g H _{2} SO}_{4} \times \dfrac{\text{1 mol H _{2} SO}_{4}}{\text{98.08 g H _{2} SO}_{4}} = \text{1.441 mol H _{2} SO}_{4}$

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ $\text{Moles of Cr _{2} (SO _{4} )}_{3} = \text{1.441 mol H _{2} SO}_{4} \times \dfrac{\text{1 mol Cr _{2} (SO _{4} )}_{3}}{\text{3 mol H _{2} SO}_{4}} = \text{0.4804 mol Cr _{2} (SO _{4} )}_{3}$

(iv) Mass of Cr₂(SO₄)₃ $\text{Mass of Cr _{2} (SO _{4} )}_{3} = \text{0.4804 mol Cr _{2} (SO _{4} )}_{3} \times \dfrac{\text{392.18 g Cr _{2} (SO _{4} )}_{3}}{\text{1 mol Cr _{2} (SO _{4} )}_{3}} = \textbf{188.4 g Cr _{2} (SO _{4} )}_{3}\\\text{The mass of Cr _{2} (SO _{4} ) _{3} formed is \large \boxed{\textbf{188.4 g}} }$

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

$\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is \large \boxed{\mathbf{98.67 \, \%}} }$