What is the mass of the precipitate formed by the interaction of a 60 g
What is the mass of the precipitate formed by the interaction of a 60 g barium chloride solution with a mass fraction of a substance of 20% with sodium sulfate?
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 60 grams of a 20% barium chloride solution.
M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol;
N BaCl2 = 60 x 0.2 / 208 = 0.058 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 0.058 = 13.51 grams;