`4.08 g` of a mixture of `BaO` and an unknown carbonate `MCO_(3)` was heated strongly. The residue weighed `3.64 g`. This was dissolved in `100 mL` of `1 N HCl`. The excess of acid required of `16 mL` of `2.5 N NaOH` for complete neutralisation. Identify the metal `M`.

Explanation

During heating `MCO_(3)` is converted into `MO` liberating `CO_(2)` while `BaO` is remaining unreacted :
`MCO_(3)(s) overset("Heat")rarr MO(s)+CO_(2)(g) uarr 0.44 g = 0.01 mol`
`(BaO(s))/(4.08 g) (BaO(s))/(3.64 g)`
From the decomposition information, it can be deduced that the original mixture contained `0.01` mole of `MCO_(3)` and the solid residue, obtained after heating contain `0.01` mole (10 millimol) of `MO`.
Also, millimol of `HCl` taken initially `= 100`
millimol of `NaOH` used in back-titration `= 16 xx 2.5 = 40`
`rArr` millimol of `HCl` reacted with oxide residue `= 60`
`HCl` reacts with oxides as :
`{:(MO +,2HCl,rarr,MCl_(2)+,H_(2)O),(10 "millimol","20 millimol",,,),(BaO+,2HCl,rarr,BaCl_(2)+,H_(2)O):}`
`60-20 = 40` millimol
Therefore, the residue contain 20 millimol of `BaO`.
Also, molar mass of `BaO = 138 + 16`
`= 154`
`rArr` Mass of `BaO = (154 xx 20)/(1000) = 3.08 g`
`rArr` Mass of `MCO_(3) = 4.08 - 3.08 = 1.0g`
`because 0.01` mole of `MCO_(3)` weight `1.0 g`
`:. 1` mole of `MCO_(3) = 100 g`
`rArr 100 = ("Atomic weight of metal") + (12 + 3 xx 16)`
`rArr` Atomic weight of metal `= 40`, i.e., `Ca`