`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 matal `M`.

Explanation

On heating, `BaO` does not decompose. Only corbonate decomposes as the follows.
`MCO_(3) rarr MO (s) + CO_(g)`
`1 mol of MCO_(3) -= 1 mol of MO`
`-= 1 mol of CO_(2)`
Los in weight on heating is due to the loss of `CO_(2)`.
Weight of `CO_(2) = 4..08 - 3.64 = 0.44 g -= 0.01 "mole" of CO_(2)`
From stochiometry,
Moles of `MO - 2 (0.01)` (`n` factor `= 2`)
Let `A =` atomic mass of `M`
Molecular mass of `MCO_(3) = a + 60`
`implies` mass of `BaO = 4.08 - 0.01 (A + 60)`
Now both the oxides, `BaO` and `MO`, will react with `HCl`. At neutralisation stage:
mEq of oxides = mEq of `HCl`
mEq of `HCl` used for oxides `= 1 xx 100 - 2.6 xx 16 = 60`
`implies mEq of MO + mEq of BaO = 60`
`implies [2(0.01) + (4.08 - 0.01 (A + 60))/(154//2)] xx 1000 = 60`
On simplifying, we get: `A = 40`
Hence, the metal `M` is calcium `(Ca)`.