Question

In a beaker, 200 mL of a 0.5M sulphuric acid solution was mixed with 300 mL of a solution of 1.2mol/L barium hydroxide. What is the mass of the precipitate formed in the reaction above?

Answer 1

The following reaction occurs:

$H_2SO_4 \ + \ Ba(OH)_2 \ = \ BaSO_4\downarrow \ + \ 2H_2O$

First need to find limiting reagent. Calculating amount of sulfuric acid:

$n(H_2SO_4)=c(H_2SO_4)V(H_2SO_4)=0.5\frac{mmol}{mL} \cdot 200 mL = 100 mmol$

Calculating amount of barium hydroxide:

$n(Ba(OH)_2)=c(Ba(OH)_2)V(Ba(OH)_2)=1.2\frac{mmol}{mL} \cdot 300 mL = 360 mmol$

Limiting reagent is sulfuric acid.

According to stoichiometry n(BaSO₄)=n(H₂SO₄)

Thus, after reaction 100 mmol of BaSO₄ will precipitate.

Mass of BaSO₄ can be found:

$m(BaSO_4)=n(BaSO_4)M(BaSO_4)=0.1 mol \cdot 233 \frac{g}{mol}= 23.3 g$