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

August 30, 2021 | education

| **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.

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