Calculate the mass of the precipitate that forms when 260 g of barium nitrate interacts with potassium sulfate.

The reaction of interaction of potassium sulfate with barium nitrate is described by the following chemical reaction equation:

Ba (NO3) 2 + K2SO4 = BaSO4 + 2KNO3;

When 1 mol of barium nitrate and 1 mol of sodium potassium sulfate interacts, 1 mol of insoluble barium sulfate is synthesized.

Let’s calculate the chemical amount of a substance that is contained in 260 grams of barium nitrate.

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

N Ba (NO3) 2 = 260/261 = 0.996 mol;

Thus, no more than 0.996 mol 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 0.996 = 232.07 grams;