A solution containing 2.61 g of potassium sulfate was added to a solution containing 2.61 g

A solution containing 2.61 g of potassium sulfate was added to a solution containing 2.61 g of barium nitrate. Calculate the mass of the precipitate formed.

According to the condition of the problem, we compose the reaction equation:
Ba (NO3) 2 + K2SO4 = BaSO4 + 2KNO3 – ion exchange reaction, barium sulfate salt is formed in the precipitate;
M Ba (NO3) 2 = 261.3 g / mol;
M (K2SO4) = 135.1 g / mol;
M (BaSO4) = 233.3 g / mol;
Determine the amount of mol of barium nitrate, potassium sulfate:
Y Ba (NO3) 2 = m / M = 2.61 / 261.3 = 0.01 mol (deficient substance);
Y (K2SO4) = m / M = 2.61 / 135.1 = 0.019 = 0.02 mol;
Calculations are made for the substance in deficiency.
Let’s make the proportion:
0.01 mol Ba (NO3) 2 – X mol (BaSO4);
-1 mol -1 mol hence, X mol (BaSO4) = 0.01 * 1/1 = 0.01 mol;
We calculate the mass of barium sulfate:
m (BaSO4) = Y * M = 0.01 * 233.3 = 2.33 g.
Answer: The mass of the barium sulfate salt is 2.33 g.