Calculate the mass of the precipitate formed by the interaction of 22 g of calcium chloride with silver nitrate.

In accordance with the condition of the problem, we compose the equation:
CaCl2 + 2AgNO3 = Ca (NO3) 2 + 2AgCl – ion exchange reaction, a precipitate of silver chloride is formed;
Let’s make the calculations:
M (CaCl2) = 111 g / mol;
M (AgCl) = 143.3 g / mol.
Determine the amount of moles of calcium chloride:
Y (CaCl2) = m / M = 22/111 = 0.198 mol.
Let’s make a proportion according to the equation:
0.198 mol (CaCl2) – X mol (AgCl);
-1 mol -2 mol hence, X mol (AgCl) = 0.198 * 2/1 = 0.396 mol.
Find the mass of AgCl salt
m (AgCl) = Y * M = 0.396 * 143.3 = 56.75 g.
Answer: The mass of silver chloride is 56.75 g.