What is the mass of the precipitate obtained by the interaction of an excess of sodium sulfate solution with 74.6 g of barium chloride solution with a mass fraction of a solute of 15%?
The reaction between sodium sulfate and barium chloride is described by the following chemical reaction equation:
BaCl2 + Na2SO4 = BaSO4 + 2NaCl;
When 1 mol of barium chloride and 1 mol of sodium sulfate interact, 1 mol of insoluble barium sulfate is synthesized.
Let’s calculate the chemical amount of a substance that is contained in 74.6 grams of a 15% barium chloride solution.
M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol;
N BaCl2 = 74.6 x 0.15 / 208 = 0.054 mol;
The same amount of barium sulfate will be synthesized.
Let’s calculate its weight.
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;
m BaSO4 = 233 x 0.054 = 12.58 grams;
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