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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. 0.226B. 0.163C. 0.307D. 0.365 |
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Answer» `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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