Mixed two solutions containing, respectively, barium chloride with a mass of 416 g

Mixed two solutions containing, respectively, barium chloride with a mass of 416 g and sodium sulfate with a mass of 426, calculate the mass of the resulting residue.

Given:
m (BaCl2) = 416 g
m (Na2SO4) = 426 g

Find:
m (draft) -?

Solution:
1) Write the reaction equation:
BaCl2 + Na2SO4 => BaSO4 ↓ + 2NaCl;
3) Calculate the amount of substance BaCl2:
n (BaCl2) = m (BaCl2) / M (BaCl2) = 416/208 = 2 mol;
5) Calculate the amount of Na2SO4 substance:
n (Na2SO4) = m (Na2SO4) / M (Na2SO4) = 426/142 = 3 mol;
6) Determine the amount of substance BaSO4:
n (BaSO4) = n (BaCl2) = 2 mol;
8) Calculate the mass of BaSO4:
m (BaSO4) = n (BaSO4) * M (BaSO4) = 2 * 233 = 466 g.

Answer: The mass of BaSO4 is 466 g.