Find the mass of the precipitate in the reaction of 71 g of sodium sulfate with barium nitrate.

1. Let’s write down the reaction of interaction of sodium sulfate with barium nitrate:

Na2SO4 + Ba (NO3) 2 = BaSO4 ↓ + 2NaNO3;

2.Calculate the chemical amount of sodium sulfate:

n (Na2SO4) = m (Na2SO4): M (Na2SO4);

M (Na2SO4) = 2 * 23 + 32 + 4 * 16 = 142 g / mol;

n (Na2SO4) = 71: 142 = 0.5 mol;

3. Determine the amount and find the mass of the precipitate of barium sulfate:

n (BaSO4) = n (Na2SO4) = 0.5 mol;

m (BaSO4) = n (BaSO4) * M (BaSO4);

M (BaSO4) = 137 + 32 + 4 * 16 = 233 g / mol;

m (BaSO4) = 0.5 * 233 = 116.5 g.

Answer: 116.5 g.