A 60 g barium hydroxide solution containing 1 percent of impurities was added to the sodium sulfate solution

A 60 g barium hydroxide solution containing 1 percent of impurities was added to the sodium sulfate solution. What is the mass of the sediment formed.

Given:
m solution (Ba (OH) 2) = 60 g
w approx. (Ba (OH) 2) = 1%
To find:
m (BaSO4)
Decision:
Ba (OH) 2 + Na2SO4 = BaSO4 + 2NaOH
w is clean. islands (Ba (OH) 2) = 100% -1% = 99% = 0.99
m in islands (Ba (OH) 2) = m solution * w pure. in-va = 60 g * 0.99 = 59.4 g
n (Ba (OH) 2) = m / M = 59.4 g / 171 g / mol = 0.35 mol
n (Ba (OH) 2): n (BaSO4) = 1: 1
n (BaSO4) = 0.35 mol
m (BaSO4) = n * M = 0.35 mol * 233 g / mol = 81.55 g
Answer: 81.55 g