What is the mass of the precipitate formed by the interaction of sulfuric
What is the mass of the precipitate formed by the interaction of sulfuric acid weighing 150 grams with a mass fraction of H2SO4 in it 49% with a solution of barium nitrate.
The reaction between sulfuric acid and barium nitrate is described by the following chemical reaction equation:
Ba (NO3) 2 + H2SO4 = BaSO4 + 2HNO3;
When 1 mol of barium nitrate and 1 mol of sodium sulfuric acid interact, 1 mol of insoluble barium sulfate is synthesized.
Let’s calculate the chemical amount of a substance that is contained in 150 grams of a 49% sulfuric acid solution.
M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;
N H2SO4 = 150 x 0.49 / 98 = 0.75 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 0.75 = 174.75 grams;