# A calcium chloride solution was added to 200 g of a silver nitrate solution until the precipitation ceased

**A calcium chloride solution was added to 200 g of a silver nitrate solution until the precipitation ceased. The mass of the resulting precipitate was 28.7 g. Calculate the mass fraction of silver nitrate in the original solution.**

Let’s execute the solution:

1. Let’s write the equation in accordance with the condition of the problem:

m = 200 g. m = 28.7 g. W -?

2AgNO3 + CaCl2 = 2AgCl + Ca (NO3) 2 – ion exchange, precipitate of chloride is released

silver;

2. Calculation of the molar masses of substances:

M (AgNO3) = 293.8 g / mol;

M (AgCl) = 143.3 g / mol.

3. Calculate the number of moles of silver nitrate, silver chloride:

Y (AgNO3) = m / M = 200 / 293.8 = 0.68 mol;

Y (AgCl) = 0.68 mol since the amount of these substances according to the equation is 1 mol.

4. Find the theoretical mass:

m (AgCl) = Y * M = 0.68 * 143.3 = 97.44 g.

5. Let’s calculate the product yield:

W = m (practical) / m (theoretical) * 100;

W = 28.7 / 97.44 = 29.45%.

Answer: the mass fraction of the silver chloride yield was 29.45%.