To a solution containing 13.6 g of barium nitrate was added 9.8 g of a sulfuric acid solution.

To a solution containing 13.6 g of barium nitrate was added 9.8 g of a sulfuric acid solution. The precipitate was filtered off and dried. What is the mass of the sediment and what substance is left in the solution?

Find the available chemical amount of barium nitrate. To do this, we divide its weight by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.

M Ba (NO3) 2 = 137 + (14 + 16 x 3) x 2 = 261 g / mol;

N Ba (NO3) 2 = 13.6 / 261 = 0.0521 mol;

Let’s calculate the available chemical amount of sulfuric acid. For this purpose, we divide its weight by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;

N H2SO4 = 9.8 / 98 = 0.1 mol;

From 0.0521 mol of barium nitrate, 0.0521 mol of barium sulfate will be obtained. In this case, sulfuric acid will remain in the solution and sodium nitrate will be added.

We calculate the weight of 0.0521 mol of barium sulfate.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;

m BaSO4 = 233 x 0.0521 = 12.1393 grams;