Question

Aluminum metal reacts with sulfuric acid according to the following equation: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(s) + 3H2( g) If, in a laboratory experiment, 19.0 g of aluminum reacts with excess sulfuric acid, and 51.1 g of Al2(SO4)3 are collected, what is the percent yield of Al2(SO4)3?

Answer 1

The percentage yield of Al₂(SO₄)₃ is 89.7114 if 19.0 g of aluminium reacts with excess sulfuric acid, and 51.1 g of Al₂(SO₄)₃ are collected.

What is a combination reaction?

A combination reaction is a reaction in which two reactants combine to form one product.

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

from the equation of reaction, 2 moles of Al react to produce 1 mole 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,

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

Solve for x

X = (19.0 g × 342.15) ÷ 54

X = 120.386g

Theoretical yield of Al₂(SO₄)₃ = 120.386g

The percentage yield of a substance = (actual yield ÷ theoretical yield) × 100

Actual yield = 108g

Theoretical yield = 120.386g

% yield = (108 ÷ 120.386g) × 100

% yield = 89.7114

The percentage yield of Al₂(SO₄)₃ is 89.7114.

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