Find the mass of the precipitate formed by the interaction of 0.01 mol of calcium chloride with sodium carbonate.

Let’s write the reaction equation:

CaCl2 + Na2CO3 = CaCO3 ↓ + 2NaCl

By condition, the amount of calcium chloride substance:

v (CaCl2) = 0.01 (mol).

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

v (CaCO3) = v (CaCl2) = 0.01 (mol).

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

m (CaCO3) = v (CaCO3) * M (CaCO3) = 0.01 * 100 = 1 (g).

Answer: 1 (g).