Hydrochloric acid was added to a solution weighing 200 g with a mass fraction of sodium silicate

Hydrochloric acid was added to a solution weighing 200 g with a mass fraction of sodium silicate of 6.1%. Calculate the mass of the precipitate formed.

Given:
m solution (Na2SiO3) = 200 g
ω (Na2SiO3) = 6.1%

To find:
m (draft) -?

Solution:
1) Na2SiO3 + 2HCl => 2NaCl + H2SiO3 ↓;
2) M (Na2SiO3) = Mr (Na2SiO3) = Ar (Na) * 2 + Ar (Si) + Ar (O) * 3 = 23 * 2 + 28 + 16 * 3 = 122 g / mol;
M (H2SiO3) = Mr (H2SiO3) = Ar (H) * 2 + Ar (Si) + Ar (O) * 3 = 1 * 2 + 28 + 16 * 3 = 78 g / mol;
3) m (Na2SiO3) = ω (Na2SiO3) * m solution (Na2SiO3) / 100% = 6.1% * 200/100% = 12.2 g;
4) n (Na2SiO3) = m (Na2SiO3) / M (Na2SiO3) = 12.2 / 122 = 0.1 mol;
5) n (H2SiO3) = n (Na2SiO3) = 0.1 mol;
6) m (H2SiO3) = n (H2SiO3) * M (H2SiO3) = 0.1 * 78 = 7.8 g.

Answer: The mass of H2SiO3 is 7.8 g.