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What will be the % loss in mass, if an e...

What will be the % loss in mass, if an equimolar mixture of `NaHCO_(3)` and `Na_(2)CO_(3)` is heated till constant weight?

A

0.226

B

0.163

C

0.307

D

0.365

Text Solution

Verified by Experts

`2NaHCO_(3)overset(Delta)rarrNa_(2)CO_(3)+H_(2)Ouparrow(g)+CO_(2)uparrow(g) n_(CO_(2))"produce"=(n_(NaHCO_(3)))/(2)`
mole a `(W_(CO_(2)))/(44)=(a)/(2)`
`Na_(2)CO_(3)(Delta)rarrx W_(CO_(2))"produce"=22a`
M.Wt. `NaHCO_(3)=84, M.Wt. Na_(2)CO_(3)=106 n_(H_(2)O)=(n_("NaHCO_(3)))/(2)`
`rArrH_(2)O` remove as vapour during heating along with `CO_(2)` give mass loss Total mass loss =22a+9a=31a
along with `CO_(2)` give mass loss
`% "loss in mass"=(31axx100)/(axx84xxaxx106)=(31)/(190)xx100=16.3%`
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