Calculate the mass of barium sulfate obtained by reacting 300 g of a 15% sulfuric acid

Calculate the mass of barium sulfate obtained by reacting 300 g of a 15% sulfuric acid solution with a sufficient amount of barium nitrate.

Given:

m (solution H2SO4) = 300 g

w% (H2SO4) = 15%

To find:

m (BaSO4) -?

Solution:

Ba (NO3) 2 + H2SO4 = BaSO4 + 2HNO3, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of sulfuric acid in the reacted solution:

m (H2SO4) = 300 g * 0.15 = 45 g

2) Find the amount of acid:

n (H2SO4) = m: M = 45 g: 98 g / mol = 0.46 mol

3) We compose a logical expression:

If 1 mol of H2SO4 gives 1 mol of BaSO4,

then 0.46 mol H2SO4 will give x mol BaSO4,

then x = 0.46 mol.

4) Find the mass of barium sulfate precipitated:

m (BaSO4) = n * M = 0.46 mol * 233 g / mol = 107.18 g.

Answer: m (BaSO4) = 107.18 g.