What is the mass of the precipitate formed when silver nitrate interacts with 40 grams of zinc chloride?

To solve the problem, let’s write down the data according to the condition:

2AgNO3 + ZnCl2 = 2AgCl + Zn (NO3) 2 – ion exchange, a precipitate of silver chloride was obtained;
Calculations:
M (ZnCl2) = 136.3 g / mol;

M (AgCl) = 143.3 g / mol;

Y (ZnCl2) = m / M = 40 / 136.6 = 0.24 mol.

Proportion:
0.24 mol (ZnCl2) – X mol (AgCl);

-1 mol -2 mol hence, X mol (AgCl) = 0.24 * 2/1 = 0.48 mol.

Find the mass of the product:
m (AgCl) = Y * M = 0.48 * 143.3 = 68.78 g

Answer: The mass of the silver chloride precipitate is 68.78 g