What mass of precipitate forms when 200.00 mL of 0.500 M NaCl is added to 650 mL of a solution that contains 15.5 g of silver sulfate?

Molar mass of Ag₂SO₄ = (107.9×2 + 32.1 + 16.0×4) g/mol = 311.9 g/mol
Molar mass of AgCl = (107.9 + 35.5) g/mol = 143.4 g/mol

2NaCl(aq) + Ag₂SO₄(aq) → 2AgCl(s) + Na₂SO₄(aq)
OR: Mole ratio NaCl(aq) : Ag₂SO₄ = 2 : 1

Initial number of moles of NaCl = (0.500 mol/L) × (200.00/1000 L) = 0.100 mol
Initial number of moles of Ag₂SO₄ = [(15.5 g) / (311.9 g/mol)] × (650/1000) = 0.0323 mol

If Ag₂SO₄ completely reacts, no. of moles NaCl needed = (0.0323 mol) × 2 = 0.0646 mol < 0.100 mol
Hence, NaCl is in excess, and thus Ag₂SO₄ completely reacts.

According to the equation, mole ratio Ag₂SO₄ : AgCl = 1 : 2
No. of moles of Ag₂SO₄ reacted = 0.0323 mol
No. of moles of AgCl precipitate formed = (0.0323 mol) × 2 = 0.0646 mol
Mass of AgCl precipitate formed = (143.4 g/mol) × (0.0646 mol) = 9.26 g (to 3 sig. fig.)