Determine the mass of the BaSO4 precipitate obtained by the interaction of 19.6gr.

Determine the mass of the BaSO4 precipitate obtained by the interaction of 19.6gr. sulfuric acid with 0.4 mol of barium hydroxide if practically the precipitate yield is 92%

Let’s execute the solution:
1. According to the condition of the problem, write the data:
H2SO4 + Ba (OH) 2 = BaSO4 + 2H2O – ion exchange, a precipitate of barium sulfate was formed;
2. Calculations by the formulas of substances:
M (H2SO4) = 98 g / mol.
M (BaSO4) = 233.3 g / mol.
Y (H2SO4) = m / M = 19.6 / 98 = 0.2 mol.
Y (BaSO4) = 0.2 mol since their number is 1 mol, which confirms the equation.
3. Find the mass of the sediment:
m (BaSO4) = Y * M = 0.2 * 233.3 = 46.7 g.
Answer: Barium sulfate weighing 46.7 g was obtained.