How many grams of calcium precipitate is formed when sodium carbonate solution is added

How many grams of calcium precipitate is formed when sodium carbonate solution is added to 250 g of 11.1% calcium chloride solution?

Let’s write the reaction equation:

CaCl2 + Na2CO3 = CaCO3 ↓ + 2NaCl

Let’s find the mass of calcium chloride:

m (CaCl2) = mr.-pa (CaCl2) * w (CaCl2) / 100 = 250 * 11.1 / 100 = 27.75 (g).

Let’s find the amount of calcium chloride substance:

v (CaCl2) = m (CaCl2) / M (CaCl2) = 27.75 / 111 = 0.25 (mol).

According to the reaction equation, from 1 mol of CaCl2, 1 mol of CaCO3 is formed, therefore:

v (CaCO3) = v (CaCl2) = 0.25 (mol).

Thus, the mass of the formed calcium carbonate precipitate is:

m (CaCO3) = v (CaCO3) * M (CaCO3) = 0.25 * 100 = 25 (g).

Answer: 25 g.