What is the mass of the precipitate formed by the interaction of 200 grams of a 30% solution of barium chloride with a solution of sodium sulfate.
ВаCl2 + Na2SO4 = 2NaCl + BaSO4.
Determine the mass of barium chloride.
m = 200 * 0.3 = 60 g.
Molar mass of barium sulfate and barium chloride.
M (BaCl2) = 208 g / mol.
M (BaSO4) = 233 g / mol.
60 g of barium chloride – x g of barium sulfate.
208 g / mol – 233 g / mol.
X = 233 * 60/208 = 67.2 g.
Answer: 67.2 g.
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