Determine the mass of the precipitate that forms when 45 grams of 60% sulfuric acid solution interacts with barium nitrate.

Given:
m solution (H2SO4) = 45 g
w (H2SO4) = 60% = 0.6
To find:
m (BaSO4)
Decision:
H2SO4 + Ba (NO3) 2 = 2HNO3 + BaSO4
m in-va (H2SO4) = m solution * w = 45 g * 0.6 = 27 g
n (H2SO4) = m / M = 27 g / 98 g / mol = 0.275 mol
n (H2SO4): n (BaSO4) = 1: 1
n (BaSO4) = 0.275 mol
m (BaSO4) = n * M = 0.275 mol * 233 g / mol = 64 g
Answer: 64 g