Question

Aluminum metal reacts with sulfuric acid according to the equation:2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(s) + 3H2(g)If 46 g of aluminum reacts with excess sulfuric acid, and 108 g of Al2(SO4)3 are collected, what is the percent yield for the reaction?

Answer 1

Answer: Percentage yield = 38.37%

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂ from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃ .

2 moles of Al contains = (2 × 27) = 54g of Al

1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃

54g of Al produces 342.15g of Al₂(SO₄)₃,

46g of Al will produce x g of Al₂(SO₄)₃

Solve for x

X = (46 × 342.15) / 54

X = 15738.9 / 54

X = 281.46g

Theoretical yield of Al₂(SO₄)₃ = 281.46g

Percentage yield of a substance = (actual yield / theoretical yield) × 100

Actual yield = 108g

Theoretical yield = 281.46

% yield = (108 / 281.46) × 100

% yield = 0.3837 × 100

%yield = 38.37

The percentage yield of Al₂(SO₄)₃ is 38.37

Answer 2

Percentage yield = 38.37%

Step-by-step explanation:

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃

2 moles of Al contains = (2 × 27) = 54g of Al

1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃

54g of Al produces 342.15g of Al₂(SO₄)₃,

46g of Al will produce x g of Al₂(SO₄)₃

Solve for x

X = (46 × 342.15) / 54

X = 15738.9 / 54

X = 281.46g

Theoretical yield of Al₂(SO₄)₃ = 281.46g

Percentage yield of a substance = (actual yield / theoretical yield) × 100

Actual yield = 108g

Theoretical yield = 281.46

% yield = (108 / 281.46) × 100

% yield = 0.3837 × 100

%yield = 38.37

The percentage yield of Al₂(SO₄)₃ is 38.37