A mixture contains equi-molar quantities of carbonates of two bivalent metals. One metal is present the extent of `13.5%` by weight in the mixture and `2.50 g` of the mixture on heating leaves a residue of `1.18 g`. Calculate the percentage by weight of the other metal.

Explanation

`2.50 g of MCO_(3)` and `XCO_(3)` (equi-molar)
`underset(x mol)(MCO_(3)) overset(Delta)rarr MO + underset(x mol)(CO_(2))`
`underset(y mol)(XCO_(3)) overset(Delta)rarr XO + underset(y mol)(CO_(2))`
Residue ` = 1.8 g`
Hence, loss in weight `= 2.5 - 1.18 = 1.32 g` (which is due to loss of `CO_(2)`)
The mixture contains metallic ions and carbonate ion. It is only carbonate ion that gives carbon dioxide. From weight of `CO_(2)`, we can calculate weight and percentage of `CO_(3)^(2-)` and hence percentage of two metals
1 mol `CO_(2) -= mol CO_(3)^(2-)`
Total moles of `CO_(2) = (1.32)/(44) = 0.03 = x + y`
= Total moles of carbonates
Total mass of `CO_(3)^(2-) = 0.03 xx 60 = 1.8 g`
`:. %` of carbonate ion `= (1.8 xx 100)/(2.5) = 72`
`:.%` of metallic ions `= 100 - 72 = 28`
`:.%` of other metal `= 28 - 13.5 = 14.5%`