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

Explanation

`BaO(xg)overset(heat)to`No change, `MCO_(3)(yg)overset(heat)to MO+CO_(2)`
`x+y=4.08` . . .[1]
Weight loss=4.08-3.64=0.44g=weight of `CO_(2)`
`CO_(2)` obtained from y g `MCO_(3)=(44y)/(M+60)`
`therefore(44y)/(M+60)=0.44` . . . [2]
millieqv. of `BaO+`millieqv. of `MCO_(3)=`millieqv. of HCl
required to neutralise BaO & MO
`=100xx1-(16xx2.5)=60`.
`(x)/(153//2)xx1000+(y)/((M+60)//2)xx1000=60` . . [3]
Solving 1,2 and 3 M=42. so, the metal is Ca.