Calculate the mass of the precipitate formed by the interaction of 35.5 g of sodium sulfate and barium nitrate.

When sodium sulfate interacts with barium nitrate, an insoluble precipitate of barium sulfate is formed. The chemical reaction is described by the following equation:
Ba (NO3) 2 + Na2SO4 = BaSO4 + 2NaNO3;
When reacted with one mole of sodium sulfate, one mole of barium sulfate is formed.
Let’s find the amount of the substance in 35.5 grams of sodium sulfate.
M Na2SO4 = 23 x 2 + 32 + 16 x 4 = 142 grams / mol;
N Na2SO4 = 35.5 / 142 = 0.25 mol;
Find the mass of 0.25 mol of barium sulfate.
Its molar mass is:
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;
The mass of insoluble salt will be:
m BaSO4 = 233 x 0.25 = 58.25 grams;