Calculate the mass (g) of barium sulfate that can be obtained by adding sulfuric acid 10% barium chloride solution to 115 g.

Given:
m solution (BaCl2) = 115 g
w (BaCl2) = 10% = 0.1
To find:
m (BaSO4)
Decision:
BaCl2 + H2SO4 = BaSO4 + 2HCl
m in-va (BaCl2) = m solution * w = 115 g * 0.1 = 11.5 g
n (BaCl2) = m / M = 11.5 g / 208 g / mol = 0.05 mol
n (BaCl2): n (BaSO4) = 1: 1
n (BaSO4) = 0.05 mol
m (BaSO4) = n * M = 0.05 mol * 233 g / mol = 11.65 g
Answer: 11.65 g