Homework Help

2.5 g of a mixture of BaO and CaO when treated with an excess of H2SO4, produced 4.713g...

user profile pic

user8235304 | Student, Grade 11 | Valedictorian

Posted June 3, 2013 at 8:14 AM via web

dislike 2 like

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.

Tagged with mole concept, science

1 Answer | Add Yours

Top Answer

user profile pic

jerichorayel | College Teacher | (Level 1) Senior Educator

Posted June 3, 2013 at 12:08 PM (Answer #1)

dislike 2 like

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`

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes