Calculate the mass of the precipitate formed by the interaction of a solution

Calculate the mass of the precipitate formed by the interaction of a solution containing sodium carbonate weighing 10.6 g with a solution containing calcium chloride weighing 5.55 g.

Given:
m (Na2CO3) = 10.6 g
m (CaCl2) = 5.55 g

To find:
m (draft) -?

1) Na2CO3 + CaCl2 => CaCO3 ↓ + 2NaCl;
2) n (Na2CO3) = m / M = 10.6 / 106 = 0.1 mol;
3) n (CaCl2) = m / M = 5.55 / 111 = 0.05 mol;
4) n (CaCO3) = n (CaCl2) = 0.05 mol;
5) m (CaCO3) = n * M = 0.05 * 100 = 5 g.

Answer: The mass of CaCO3 is 5 g.