Calculate the mass of the precipitate that will precipitate when an excess of potassium carbonate interacts with

Calculate the mass of the precipitate that will precipitate when an excess of potassium carbonate interacts with 17.4 g of a solution of barium nitrate with a mass fraction of the latter of 15%

Given:
m solution (Ba (NO3) 2) = 17.4 g
ω (Ba (NO3) 2) = 15%

To find:
m (draft) -?

1) K2CO3 + Ba (NO3) 2 => 2KNO3 + BaCO3 ↓;
2) m (Ba (NO3) 2) = ω * m solution / 100% = 15% * 17.4 / 100% = 2.61 g;
3) n (Ba (NO3) 2) = m / M = 2.61 / 261 = 0.01 mol;
4) n (BaCO3) = n (Ba (NO3) 2) = 0.01 mol;
5) m (BaCO3) = n * M = 0.01 * 197 = 1.97 g.

Answer: The mass of BaCO3 is 1.97 g.