What is the theoretical yield of aluminum sulfate ? (grams) What is the percent yield of aluminum sulfate ? (%)

Question

For the following reaction, 5.37 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.7 grams of aluminum sulfate. aluminum oxide (s) + sulfuric acid (aq) aluminum sulfate (aq) + water (l)

Answer 1

Al₂O₃(s) + 3H₂SO₄(aq) = Al₂(SO₄)₃(aq) + 3H₂O(l)

molar mass of Al₂O₃ = (27 * 2) + (16 * 3) = 102 g/mol

molar mass of Al₂(SO₄)₃ = (27 * 2) + (3 * (32 + (16 * 4))) = 342 g/mol

number of moles = mass/molar mass

number of moles of Al₂O₃ = 5.37/102 = 0.053 mol

According to the balanced reaction, 1 mol of Al₂O₃ produces 1 mol of Al₂(SO₄)₃.

So, 0.053 moles of Al₂O₃ produce 0.053 moles of Al₂(SO₄)₃.

theoretical yield of Al₂(SO₄)₃ = 0.053 mol * 342 g/mol = 18.126 g

actual yield of Al₂(SO₄)₃ = 12.7 g

percent yield of Al₂(SO₄)₃ = 12.7/18.126 * 100% = 70.07%