Determine the mass of the precipitate that is formed by draining 15 g of a 5% BaCl2 solution and 10 g of an 8% Na2So4 solution.

Given:
m solution (BaCl) = 15 g
w (BaCl) = 5% = 0.05
m solution (Na2SO4) = 10 g
w (Na2SO4) = 8% = 0.08
To find:
m (BaSO4)
Decision:
BaCl2 + Na2SO4 = BaSO4 + 2NaCl
m in-va (BaCl2) = w * m solution = 0.05 * 15 g = 0.75 g
m in-va (Na2SO4) = w * m solution = 0.08 * 10 g = 0.8 g
n (BaCl2) = m / M = 0.75 g / 208 g / mol = 0.004 mol
n (Na2SO4) = m / M = 0.8 g / 142 g / mol = 0.006 mol
Na2SO4 in g. then
n (BaCl2): n (BaSO4) = 1: 1
n (BaSO4) = 0.004 mol
m (BaSO4) = n * M = 0.004 mol * 233 g / mol = 0.932 g
Answer: 0.932 g