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2.5 g of a mixture of BaO and CaO when treated with an excess of H2SO4,...
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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`
`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.
Posted by jeew-m on June 28, 2013 at 8:35 AM (Answer #1)
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