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.

Expert Answers
jerichorayel eNotes educator| Certified Educator

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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