Calculate the mass of precipitate formed when 200 ml of 0.5 M Silver Nitrate reacts with excess Sodium Chloride.
- 14.35 g
- 10.8 g
- 7.17 g
- 28.7 g
📝 Detailed explanation
AgNO₃ + NaCl → AgCl + NaNO₃.n(AgNO₃) = 0.2 * 0.5 = 0.1 mol.
n(AgCl) = 0.1 mol.
M(AgCl) = 108 + 35.5 = 143.5 g/mol.
Mass = 0.1 * 143.5 = 14.35 g.