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

Question

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

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

Monjur Alahi

11 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`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%`