Calculate the mass of the precipitate that will fall out when 100 g of 11% CaCl2 solution interacts with AgNO3.

Given:
m solution (CaCl2) = 100 g
ω (CaCl2) = 11%

To find:
m (draft) -?

1) CaCl2 + 2AgNO3 => Ca (NO3) 2 + 2AgCl ↓;
2) m (CaCl2) = ω (CaCl2) * m solution (CaCl2) / 100% = 11% * 100/100% = 11 g;
3) M (CaCl2) = Mr (CaCl2) = Ar (Ca) * N (Ca) + Ar (Cl) * N (Cl) = 40 * 1 + 35.5 * 2 = 111 g / mol;
4) n (CaCl2) = m (CaCl2) / M (CaCl2) = 11/111 = 0.1 mol;
5) n (AgCl) = n (CaCl2) * 2 = 0.1 * 2 = 0.2 mol;
6) M (AgCl) = Mr (AgCl) = Ar (Ag) * N (Ag) + Ar (Cl) * N (Cl) = 108 * 1 + 35.5 * 1 = 143.5 g / mol;
7) m (AgCl) = n (AgCl) * M (AgCl) = 0.2 * 143.5 = 28.7 g.

Answer: The mass of AgCl is 28.7 g.