Answer to Question #90485 in General Chemistry for acadia carlson

Question #90485
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?
1
Expert's answer
2019-06-03T09:17:16-0400

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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