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

1 Answer

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`BaO+H_2SO_4 rarr BaSO_4+H_2O`

`CaO+H_2SO_4 rarr CaSO_4+H_2O`


Molar mass in g/(mol)

`BaO = 153`

`CaO = 56`

`CaSO_4 = 136`

`BaSO_4 = 233`


Mole ratio

`BaO:BaSO_4 = 1:1`

`CaO:CaSO_4 = 1:1`


Let us say we have xg of BaO in the mixture. So we should have (2.5-x)g of CaO.

Amount of BaO moles` = x/153`

Amount of CaO moles` = (2.5-x)/56`


Since mole ratio is `1:1` ;

Amount of `BaSO_4 ` moles `= x/153`

Amount of `CaSO_4` moles `= (2.5-x)/56`


Weight of `BaSO_4 = x/153xx233`

Weight of `CasO_4 = (2.5-x)/56xx136`


But the weight of the sulphate mixture is 4.713g

`x/153xx233+(2.5-x)/56xx136 = 4.713`

`x = 1.5`


So the weight of BaO is 1.5g.

Percentage weight of BaO `= 1.5/2.5xx100% = 60%`


So we have 60% of BaO in the mixture.



The mixture only contains BaO and CaO without any impurities.