How many grams of a 10% solution of calcium chloride will enter into an exchange reaction with a solution

How many grams of a 10% solution of calcium chloride will enter into an exchange reaction with a solution of sodium carbonate, if as a result 0.3 mol of a precipitate is formed.

Given:
w (CaCl2) = 10% = 0.1
n (CaCO3) = 0.3 mol
To find:
m solution (CaCl2)
Decision:
CaCl2 + Na2CO3 = 2NaCl + CaCO3
n (CaCO3): n (CaCl2) = 1: 1
n (CaCl2) = 0.3 mol
m in islands (CaCl2) = n * M = 0.3 mol * 111 g / mol = 33.3 g
m solution (CaCl2) = m in-va / w = 33.3 g / 0.1 = 333 g
Answer: 333 g