What mass of precipitate is formed when a solution of barium chloride interacts with sulfuric acid
What mass of precipitate is formed when a solution of barium chloride interacts with sulfuric acid, the mass of which is 160 g, with W (sulfuric acid) = 15 percent.
The reaction between sulfuric acid and barium chloride is described by the following chemical reaction equation:
BaCl2 + H2SO4 = BaSO4 + 2HCl;
In the interaction of 1 mol of barium chloride and 1 mol of sulfuric acid, 1 mol of insoluble barium sulfate is synthesized.
Let’s calculate the chemical amount of a substance that is contained in 160 grams of a 15% sulfuric acid solution.
M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;
N H2SO4 = 160 x 0.15 / 98 = 0.245 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 0.245 = 57.08 grams;