If 49.2 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams of Ag₂SO₄ that could be formed? AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)

Answer 1

AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)

Balanced: 2 AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2 HNO₃ (aq)

AgNO₃ is 169.87 g/mol, so 49.2 g is 0.290 mol.

H₂SO₄ is 98.08 g/mol, so 28.6 g is 0.292 mol.

Because the ratio of AgNO₃ to H₂SO₄ is 2:1, we would need twice the moles of AgNO₃ as H₂SO₄. This makes AgNO₃ our limiting reagent. We would use 0.290 mol of AgNO₃ and 0.145 mol of H₂SO₄.

Our ratio of AgNO₃ to Ag₂SO₄ is also 2:1, so we would produce half the moles of Ag₂SO₄. We would use 0.290 mol of AgNO₃ and produce 0.145 mol of Ag₂SO₄.

Ag₂SO₄ is 311.80 g/mol, so 0.145 mol of Ag₂SO₄ is 45.211 g.