How many grams of CaCO3 will precipitate if an excess of soda solution is added to 400 ml of 0.5 N CaCi2 solution?

Given:
V solution (CaCl2) = 400 ml = 0.4 l
Cn (CaCl2) = 0.5 n

To find:
m (CaCO3) -?

Decision:
1) Write the equation of a chemical reaction:
CaCl2 + Na2CO3 => CaCO3 ↓ + 2NaCl;
2) Calculate the amount of CaCl2:
n (CaCl2) = CH (CaCl2) * V solution (CaCl2) / z = 0.5 * 0.4 / 2 = 0.1 mol;
3) Determine the amount of CaCO3 substance (taking into account the coefficients in the reaction equation):
n (CaCO3) = n (CaCl2) = 0.1 mol;
4) Calculate the mass of CaCO3:
m (CaCO3) = n (CaCO3) * M (CaCO3) = 0.1 * 100 = 10 g.

Answer: The mass of CaCO3 is 10 g.