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How much propanol is required for dehydration to get 2.24 litre of propene at N.T.P. If yield is 100% |
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Answer» Solution :`C_(3)H_(8)O+H_(2)SO_(4)rarrC_(3)H_(6)+H_(2)O+H_(2)SO_(4)` Molecular weight of propanol `=60` from the equation given above we can see that from dehydration of `1` mole or `60` GRAM of propanol we get `1` mole (`22.4`lit) of PROPENE as PRODUCT. `because 22.4` litre of `C_(3)H_(6)` can be get from dehydration of `60g` of propanol. `therefore 1` litre of propene can be get from dehydration of `(60)/(22.4)g` of propanol |
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