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`)

Answer

`00009.25`

Explanation

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