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

Explanation

BaO does not change on heating
Suppose the weight of `MCO_(3)` is x g and at . Wt of M is y
`{:(MCO_(3) to ,MO +,CO_(2)),(" x g",[3.64 -(4.08-x)],0.44 g ),(,=(x - 0.44)g,):}`
Applying POAC for m and C atoms, we get
`x/(y+60) = ((x-0.44))/(y+16) " "...(1)`
and `x/(y+60) = (0.44)/44 = 0.01 " "...(2)`
From eqns . (1) and (2) , we have ,
` (x-0.44)/(y+16) = 0.01 " "...(3)`
Now , m.e of NaOh = `2.5 xx16 = 40 " "...(Eqn.1)`
` :.` m.e of excess acid = 40
` :. ` m.e of the acid used to neutralise BaO and MO
= m.r of total acid = m.e of excess acid
` = 1xx 100 - 40 = 60`
` :. ` eq. of the acid = `60/100 = 0.06` = eq. of BaO + eq. of MO
or `(4.08 -x)/(154//2) + ((x-0.44))/((y+16)2) = 0.06`
`("eq. wt . of BaO "= 154/2 , " eq. wt of MO "= (y+16)/2)`
Substituting the value of `((x-0.44)/(y+16)) ` from Eqn . (3) in Eqn .(4)
we get , x = 1 and y = 40
hence , the metal M must be Ca.