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"
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