# When a 16 g sulfuric acid solution interacted with an excess of barium chloride solution, a 5.7 g

**When a 16 g sulfuric acid solution interacted with an excess of barium chloride solution, a 5.7 g precipitate was formed. Determine the mass fraction of sulfuric acid in the original solution.**

1. Let’s compose the equation of the proceeding reaction:

BaCl2 + H2SO4 = BaSO4 ↓ + 2HCl;

2. Calculate the chemical amount of barium sulfate:

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

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

n (BaSO4) = 5.7: 233 = 0.0245 mol;

3. Set the amount of reacted sulfuric acid:

n (H2SO4) = n (BaSO4) = 0.0245 mol;

4. Find the mass of sulfuric acid:

m (H2SO4) = n (H2SO4) * M (H2SO4);

M (H2SO4) = 2 + 32 + 4 * 16 = 98 g / mol;

m (H2SO4) = 0.0245 * 98 = 2.4 g;

5. Determine the mass fraction of sulfuric acid in the original solution:

w (H2SO4) = m (H2SO4): m (solution) = 2.4: 16 = 0.15 or 15%.

Answer: 15%.