What is the mass of the precipitate precipitated when 196 g of a 20% solution of H2SO4 interacts with an excess of (BaNO3) 2?

H2SO4 + Ba (NO3) 2 = 2HNO3 + BaSO4

m H2SO4 solution = 196 g

w% (percentage of sulfuric acid in this solution) = 20%

Then the mass of pure sulfuric acid:

m = m solution * w%.

The amount of sulfuric acid:

n = m / M, where M is the molar mass of sulfuric acid = 98 g / mol.

According to the reaction equation, we see that the stoichiometric coefficients of sulfuric acid and barium sulfate are equal (= 1), which means their number is equal.

Barium sulfate mass: m (baso4) = n * M (baso4) = [m solution * w% / M (H2SO4)] * M (BaSO4) = [196 * 0.2 / 98] * 233 = 93.2 g.

Answer: m (BaSO4) = 93.2 g.

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