14 December, 06:39

# 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?

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1. 14 December, 07:41
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Percentage yield = 38.37%

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