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A mixture of `NaCl` and `Na_(2)CO_(3)` 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.A. `6.6` gB. `5.8` gC. `6.8` gD. `7.2` g |
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Answer» Correct Answer - A `CO_(2)` is from `Na_(2) CO_(3)` only `underset(underset((?))(106g))(Na_(2)CO_(3)) + 2HCl rarr2 NaCl + underset(underset(2,24 g)(44 g)) (CO_(2))+ H_(2) O ` 44 g `CO_(2)` is from = 106 g `Na_(2) CO_(3)` `therefore 2.24 g CO_(2) " is from " = 106/44 xx 2.24 " g " Na_(2) CO_(3) = 5.4g` Thus, `NaCl = 12 - 5.4 = 6.6` g |
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