Answer:
[tex]\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}[/tex]
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
a) Mass of Cr₂(SO₄)₃
(i) Mass of pure H₂SO₄
[tex]\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}[/tex]
(ii) Moles of H₂SO₄
[tex]\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}[/tex]
(iii) Moles of Cr₂(SO₄)₃
The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ [tex]\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}[/tex]
(iv) Mass of Cr₂(SO₄)₃ [tex]\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 $\large \boxed{\textbf{188.4 g}}$}[/tex]
b) Percentage yield
It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.
[tex]\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 $\large \boxed{\mathbf{98.67 \, \%}}$}[/tex]
