An excess of barium hydroxide solution was added to 40 grams of 12% sulfuric acid solution.

An excess of barium hydroxide solution was added to 40 grams of 12% sulfuric acid solution. Determine the mass of the precipitate formed.

Given:

m (H2SO4) = 40 g

w% (H2SO4) = 12%

Find:

m (draft) -?

Solution:

1) We compose the reaction equation typical for this problem:

H2SO4 + Ba (OH) 2 = 2H2O + BaSO4;

2) Find the mass of acid in solution:

m (H2SO4) = 40 g * 0.12 = 4.8 g;

3) Find the amount of sulfuric acid:

n (H2SO4) = m: M = 4.8 g: 98 g / mol = 0.049 mol;

4) We compose logical equality:

if 1 mol of H2SO4 gives 1 mol of BaSO4,

then 0.049 mol H2SO4 will give x mol BaSO4,

then x = 0.049 mol.

5) Find the mass of precipitated barium sulfate:

m (BaSO4) = n * M = 0.049 mol * 233 g / mol = 11.417 g;

Answer: m (BaSO4) = 11.417 g.