What is the amount of precipitate released during the interaction of sodium sulfate with barium

What is the amount of precipitate released during the interaction of sodium sulfate with barium chloride if the mass of sulfate is 25 g?

Given:
m (Na2SO4) = 25 g

To find:
n (sediment) -?

Decision:
1) Write the equation of a chemical reaction:
Na2SO4 + BaCl2 => BaSO4 ↓ + 2NaCl;
2) Calculate the molar mass of Na2SO4:
M (Na2SO4) = Mr (Na2SO4) = Ar (Na) * N (Na) + Ar (S) * N (S) + Ar (O) * N (O) = 23 * 2 + 32 * 1 + 16 * 4 = 142 g / mol;
3) Calculate the amount of Na2SO4 substance:
n (Na2SO4) = m (Na2SO4) / M (Na2SO4) = 25/142 = 0.18 mol;
4) Determine the amount of substance BaSO4:
n (BaSO4) = n (Na2SO4) = 0.18 mol.

Answer: The amount of substance BaSO4 is 0.18 mol.