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When `78.5gm` of `CH_(3)COCl` is treated with `29.75gm` of `CH_(3)MgBr` then what will be the mass of alcohol produces? (Given: `M`. wt. of `Cl=35.5, Br=80, C=12`) |
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Answer» Correct Answer - `00009.25` Moles of `CH_(3)COCL=78.5/78.5=1.0` mole Moles of `CH_(3)MgBr=29.75/119=0.25` `:. CH_(3)MgBr` is limiting reagent `H_(3)C-overset(O)overset(||)(C)-Cl+2CH_(3)MgBrtoH_(3)C-underset(CH_(3)) underset(|) overset(OH) overset(|)(C)-CH_(3)` Moles of alcohol `=0.25/2` Weight of alcohol `=0.25/2xx74=9.25` gm |
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