4.08g ofa mixtureof 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.

4.08g ofa mixtureof BaO and an unkonwn carbonate MCO_(3) was heated st

1.	4.08g ofa mixtureof 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.
Answer» Solution : BAO does not change on heating Suppose the weightof `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.