Answer:
a. 167 mL b. 39.27 %
Explanation:
a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol Al₂(SO₄)₃.
Now, we calculate the number of moles of Al in 45.0 g Al.
We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.
So n = 45.0 g/26.98 g/mol = 1.668 mol
Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol
So, we have 2.5 mol H₂SO₄.
Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.
V = n/C
= 2.5 mol/15.0 mol/L
= 0.167 L
= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.
b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃
Therefore 1.668 mol Al will produce x mol Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol
So, we need to find the mass of 0.834 mol Al₂(SO₄)₃. Now molar mass Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.
Also number of moles of Al₂(SO₄)₃, n = mass of Al₂(SO₄)₃,m/molar mass Al₂(SO₄)₃, M
n =m/M
So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g
% yield = Actual yield/theoretical yield × 100 %
Actual yield = 112 g, /theoretical yield = 285.2 g
So, % yield = 112 g/285.2 g × 100 %
= 0.3927 × 100 %
= 39.27 %
