Explanation:
Aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas according to the following equation.
2 Al (s) + 3 H₂SO₄ (aq) ----> Al₂(SO₄)₃ (aq) + 3 H₂ (g)
We have to find the mass of Al₂(SO₄)₃ that would be formed if 88.5 g of H₂SO₄ completely reacted with Al.
First we have to convert the mass of sulfuric acid into moles using its molar mass.
molar mass of H₂SO₄ = 98.08 g/mol
moles of H₂SO₄ = 88.5 g * 1 mol/(98.08 g)
moles of H₂SO₄ = 0.902 moles
2 Al (s) + 3 H₂SO₄ (aq) ----> Al₂(SO₄)₃ (aq) + 3 H₂ (g)
Now, if we pay attention to the coefficients of the reaction we will see that 2 moles of Al will react with 3 moles of H₂SO₄ to produce 1 mol of Al₂(SO₄)₃ and 3 moles of H₂. So the molar ratio between H₂SO₄ and Al₂(SO₄)₃ is 3 to 1. We can use that relationship to find the number of moles of Al₂(SO₄)₃ that can be produced when 0.902 moles of H₂SO₄ reacted with excess Al.
3 moles of H₂SO₄ : 1 mol of Al₂(SO₄)₃ mole ratio
moles of Al₂(SO₄)₃ = 0.902 moles of H₂SO₄ * 1 mol of Al₂(SO₄)₃/(3 moles of H₂SO₄)
moles of Al₂(SO₄)₃ = 0.300 moles
And finally we can convert those moles back to grams using the molar mass of Al₂(SO₄)₃.
molar mass of Al₂(SO₄)₃ = 342.15 g/mol
mass of Al₂(SO₄)₃ = 0.300 moles * 342.15 g/mol
mass of Al₂(SO₄)₃ = 102.6 g
Answer: b. 103 g