1.

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 is

Answer»

`56 g`
`28g`
`42G`
`20g`

Solution :`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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