# Determine the mass of the precipitate that was obtained by the interaction of 23 g of calcium chloride with

**Determine the mass of the precipitate that was obtained by the interaction of 23 g of calcium chloride with sodium carbonate? If the practical sludge yield is 85%**

Let’s implement the solution:

In accordance with the condition, we will write the data:

CaCl2 + Na2CO3 = CaCO3 + 2NaCl – ion exchange, calcium carbonate precipitate is formed;

Let’s make calculations using the formulas:

M (CaCl2) = 111 g / mol;

M (CaCO3) = 100 g / mol;

Y (CaCl2) = m / M = 23/111 = 0.2 mol;

Y (CaCO3) = 0.2 mol since the amount of these substances is 0.2 mol.

We find the mass of the sediment, taking into account the product yield:

m (CaCO3) = Y * M = 0.2 * 100 = 20 g (theoretical weight);

m (CaCO3) = 0.85 * 20 = 17 g (practical weight).

Answer: Received calcium carbonate weighing 17 g