If 6.79 g of aluminum were reacted with excess sulfuric acid to produce 37.86 g of aluminum sulfate, what is the percent yield?

Molar mass of Al = 27.0 g/mol
Molar mass of Al₂(SO₄)₃ = (27.0×2 + 32.1×3 + 16.0×12) g/mol = 342.3 g/mol

2Al(l) + 3H₂SO₄(aq) → Al₂(SO₄)₃ + 3H₂(aq)
Mole ratio Al : Al₂(SO₄)₃ = 2 : 1

No. of moles Al reacted = (6.79 g) / (27.0 g/mol) = 0.2515 mol
Maximum no. of moles of Al₂(SO₄)₃ produced = (0.2515 mol) × (1/2) = 0.1258 mol
Theoretical yield of Al₂(SO₄)₃ = (0.1258 mol) × (342.3 g/mol) = 43.06 g

Percent yield = (37.86/43.06) × 100(%) = 87.9(%)

2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2

(6.79 g Al) / (26.98154 g Al/mol) x (1 mol Al2(SO4)3 / 2 mol Al) x (342.1509 g Al2(SO4)3/mol) =
43.0517 g = 43.1 g Al2(SO4)3 in theory

(37.86 g) / (43.0517 g) = 0.8794 = 87.94% yield Al2(SO4)3