Find the mass fraction of the yield of barium sulfate if, upon the interaction of 17.1 g
Find the mass fraction of the yield of barium sulfate if, upon the interaction of 17.1 g of barium hydroxide with sulfuric acid, 20 g of barium sulfate were obtained
Barium hydroxide reacts with sulfuric acid. This results in the formation of water-insoluble barium sulfate, which precipitates. This reaction is described by the following chemical equation.
Ba (OH) 2 + H2SO4 = BaSO4 + 2H2O;
Barium hydroxide reacts with sulfuric acid in equivalent molar amounts. In this case, an equal amount of insoluble salt is synthesized.
Let’s calculate the chemical amount of barium hydroxide.
M Ba (OH) 2 = 137 + 16 x 2 + 2 = 171 grams / mol; N Ba (OH) 2 = 17.1 / 171 = 0.1 mol;
Barium sulfate will be synthesized in the same molar amount.
Let’s calculate its weight.
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.1 x 233 = 23.3 grams;
The reaction yield will be 20 / 23.3 = 0.858 = 85.8%;