An excess of sodium carbonate solution was added to 200 g of barite water, and a precipitate weighing 8.9 g precipitated. Determine the mass fraction of barium hydroxide in the initial solution.
1) Let’s compose the reaction equation:
Ba (OH) 2 + Na2CO3 = BaCO3 ↓ + 2NaOH
2) Calculate the amount of barium carbonate and barium hydroxide substance:
n (BaCO3) = m (BaCO3) / M (BaCO3) = 8.9 / 197 = 0.045 mol.
according to the reaction equation
n (Ba (OH) 2) = n (BaCO3) = 0.045 mol
3) Determine the mass of barium hydroxide and its mass fraction in barite water:
m (Ba (OH) 2) = n (Ba (OH) 2) ∙ M (BaCO3) = 0.045 ∙ 171 = 7.695 g.
ω (Ba (OH) 2) = (m (Ba (OH) 2) / m (solution)) ∙ 100% = (7.695 / 200) ∙ 100 = 3.85%
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