Calculate the mass of silver chloride, which is formed by dissolving 116 g of silver nitrate containing 3%

Calculate the mass of silver chloride, which is formed by dissolving 116 g of silver nitrate containing 3% impurities in an excess of hydrochloric acid.

1. AgNO3 + HCl = AgCl ↓ + HNO3;

2.w (AgNO3) = 100% – 3% = 97% or 0.97 shares;

3. find the mass of silver nitrate without impurities:

m (AgNO3) = w (AgNO3) * m (nitrate with impurities);

m (AgNO3) = 0.97 * 116 = 112.52 g;

4.n (AgNO3) = m (AgNO3): M (AgNO3);

M (AgNO3) = 108 + 14 + 16 * 3 = 170 g / mol;

n (AgNO3) = 112.52: 170 = 0.6619 mol;

5.n (AgCl) = n (AgNO3) = 0.6619 mol;

6. find the mass of silver chloride:

m (AgCl) = n (AgCl) * M (AgCl);

M (AgCl) = 108 + 35.5 = 143.5 g / mol;

m (AgCl) = 0.6619 * 143.5 = 94.98 g.

Answer: 94.98 g.