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A mixture of `NaCl` and `Na_(2)CO` is given On heating `12 g` of the mixture with dilute `HCl, 2.24 g` of `CO_(2)` is removed. Calculate the amounts of each in the mixture. |
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Answer» Reactions: `NaCl + HCl rarr` no reaction `Na_(2) CO_(3) + 2 HCl rarr 2 NaCl + CO_(2) + H_(2) O` Let `x g` of `Na_(2)CO_(3)` in the mixture. `:. (x)/(106) mol of Na_(2) CO_(3) = (x)/(106) mol of CO_(2)` `:. (x)/(106) = (2.24)/(44)` or `x = 5.4 g Na_(2) CO_(3)` Weight of `NaCl` in mixture `= 12 - 5.4 = 6.6 g` |
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