What is the mass of the precipitate formed by the interaction of barium nitrate with a solution
What is the mass of the precipitate formed by the interaction of barium nitrate with a solution of sulfuric acid weighing 34.8 g with a mass fraction of 5%?
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 the substance, which is contained in 34.8 grams of 5% sulfuric acid solution.
M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;
N H2SO4 = 34.8 x 0.05 / 98 = 0.018 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 0.018 = 4.19 grams;