2.5 g of a mixture of BaO and CaO when treated with an excess of H2SO4, produced 4.713g of the mixed sulphates. Find the percentage of BaO present in the mixture.

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2.5 g = mass BaO + Mass CaO

Let mass BaO = x; mass Cao = y

`2.5 g = x + y` **equation 1**

4.713 g = mass BaSO4 + mass CaSO4

By stoichiometry:

`4.713 g = x (MW BaSO4)/(MW BaO) + y (MW CaSO4)/(MW CaO)` **equation 2**

We know that:

`2.5 g = x + y`

`y = 2.5 - x`

Substitute equation 2 in the equation 1

`4.713 g = x (MW BaSO4)/(MW BaO) + y (MW CaSO4)/(MW CaO)`

`4.713 g = x (MW BaSO4)/(MW BaO) + 2.5 - x (MW CaSO4)/(MW CaO)`

Molar weights, MW (g/mol):

BaO = 153.33

BaSO4 = 233.39

CaO = 56.08

CaSO4 = 136.14

`4.713 = x (233.39)/(153.33) + 2.5 - x (136.14)/(56.08)`

`4.713 = 1.522x + 2.5 - x (2.428)`

`4.713 = 1.522x + 6.07 - 2.428x`

`2.428x - 1.522x = 6.07 - 4.713`

`0.906x = 1.357`

`x = 1.498 grams BaO`

`%BaO = 1.498/2.5 * 100 = 60% -> answer`

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