What is the mass of the precipitate formed when the solutions containing 35 grams of potassium chloride

What is the mass of the precipitate formed when the solutions containing 35 grams of potassium chloride and 50 grams of silver nitrate are poured?

1. We make a chemical equation, not forgetting to put down the coefficients:

KCl + AgNO3 = AgCl + KNO3.

2. Find the amount of substance KCl:

n (KCl) = m / M = 35 / 74.5 = 0.469 mol.

3. Let’s find the amount of substance AgNO3:

n (AgNO3) = m / M = 50/170 = 0.294 mol.

4. We see that:

n (AgNO3) <n (KCl); Therefore, further calculations will be carried out using AgCl (silver nitrate).

5. Find the amount of AgCl:

n (AgCl) = n (AgNO3) = 0.294 mol.

6. Find the mass of AgCl:

m (AgCl) = n * M = 0.294 * 143.5 = 42.189g.

Answer: m (AgCl) = 42.189g.