`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

Weight of mixture of BaO and `MCO_3=4.08g`
On heating of `MCO_3` gives MO and `CO_2`
Weight of `BaO+MO=3.64g`
Weight of `CO_2` evolved `=4.08-3.64=0.44g`
Mols of `CO_2=(0.44)/(44)=10^(-2)mol=10^(-2)xx10^(3)`
`=10m mol =20 m" Eq of "CO_2`
`=20mEq MCO_3`
Total acid `=100xx1=100mEq`
Excess of acid`=`Total acid`-`m" Eq of "`NaOH`
`=100-16xx2.5=60mEq` of excess acid
`(Ew(BaO)=(Mw)/(2))`
`m" Eq of "BaO=60-20=40` mEq
`=((138+16)xx40)/(100xx2)=3.08g`
weight of `MO=3.64-3.08=0.56g`
20 m" Eq of "`MO=0.56`
Equivalent weight of `MO=0.56xx1000=28`
Molecular weight of `MO=28xx2=56`
Hence, atomic weight of `M=56-16=40`. The metal is