The interaction of barium nitrate with sulfuric acid resulted in a precipitate weighing 139.8 g.
The interaction of barium nitrate with sulfuric acid resulted in a precipitate weighing 139.8 g. Calculate the mass of the sulfuric acid that has reacted.
Barium nitrate reacts with sulfuric acid. In this case, a water-insoluble barium sulfate salt is formed, which turns into an insoluble precipitate. The reaction is described by the following chemical reaction equation.
Ba (NO3) 2 + H2SO4 = BaSO4 + 2HNO3;
Barium nitrate reacts with sulfuric acid in equal molar amounts. In this case, the same amount of insoluble barium sulfate is synthesized.
Determine the chemical amount of barium sulfate.
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; N BaSO4 = 139.8 / 233 = 0.6 mol;
The same amount of sulfuric acid will be used.
Let’s find its weight.
M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; m H2SO4 = 0.6 x 98 = 58.8 grams