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The decomposition of a certain mass of `CaCO_(3)` gave `11.2 dm^(3)` of a `CO_(2)` at S.T.P. The mass of KOH required to completely neutralize the gas isA. `56 g`B. `28g`C. `42g`D. `20g` |
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Answer» Correct Answer - A `underset(112g)underset(2(39+16+1))(2KOH)+underset("at S.T.P.")underset(22.4 L)(CO_(2))toK_(2)CO_(3)+H_(2)O` `22.4 dm^(3)` of `CO_(2)` at N.T.P. required KOH = 112 g `11.2 dm^(3)` of `CO_(2)` at S.T.P. will require KOH = 56 g |
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