Question: > 0.05 mole of LiAlH4 in ether solution was placed in a flask containing 74g (1 mole) of t-butyl alc...

Question

0.05 mole of LiAlH4 in ether solution was placed in a flask containing 74g (1 mole) of t-butyl alcohol. The product LiAlHC12H27O3 weighed 12.7 g. If Li atoms are conserved, the percentage yield is : (Li = 7, Al = 27, H = 1, C = 12, O = 16).

Answer 1

Solution

(3)

Li AlH4 + t-butyl alcohol $\overset{\quad Ether\quad}{\rightarrow}$ (M.W. = 254)

0.05 mole 12.7 gram

$= \frac{12.7}{254}$ = 0.05 mole

Li atom remain conserved so

No. of mole of LiAlH4 = No. of mole of LiAlHC12H27O3

So No. of mole of LiAlHC12H27O3 = 0.05

% yield $= \frac{0.05}{0.05} \times 100 = 100\%$