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InterviewSolution
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Two moles of HCHO and 1 mol of PhCHO react with conc. NaOH. What are the produsts quantitatively? |
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Answer» Initially there will be crossed Cannizzaro reaction between HCHO and PhCHO. `HCHO+PhCHO overset(NaOH)rarr underset(1mol)(HCOONa) + underset(1mol)(PhCH_(2)OH)` `underset((1/2)mol)(HCHO) + underset((1/2)mol)(HCHO) overset(NaOH)rarr underset((1/2)mol)(HCOONa) + underset((1/2)mol)(CH_(3)OH)` Remaining 1 mol of HCHO will give Cannizzaro reaction. So `1//2` mol of HCHO will react with `1//2` mol of HCHO to give `1//2` mol of HCOONa and `1//2` mol of `CH_(3)OH`. Hence, `1.5` mol of HCOONa, 1 mol of `PhCH_(2)OH`, and `1//2` mol of `CH_(3)OH` are formed. |
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