A solution of 11.92 g of potassium chloride was added to a solution containing 25.5 g of silver nitrate.

A solution of 11.92 g of potassium chloride was added to a solution containing 25.5 g of silver nitrate. Determine the mass of the sediment formed

AgNO3 + KCl = KNO3 + AgCl (down.)
The problem for excess and lack:
1) n (AgNO3) = 25.5 / 170 = 0.15 mol.
2) n (KCl) = 11.92 / 74.5 = 0.16 mol.
3) By condition: n (AgNO3) / n (KCl) = 0.15 / 0.16 = 0.9375
Equation: n (AgNO3) / n (KCl) = 1/1 = 1
It turns out, therefore, KCl is in excess, we calculate by AgNO3:
4) n (AgCl) = n (AgNO3) = 0.15 mol.
5) m (AgCl) = 0.15 * 143.5 = 21.525 g.
Answer: m (AgCl) = 21.525 g.