`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`.

`4.08 g` of a mixture of `BaO` and an unknown carbonate `MCO_(3)` was

1.	`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`.
Answer» 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

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`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`.

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