0.05 mole of LiAlH4 in either solution was placed in a flask contaning 74g (1mole) of t-butyl alcohol. The product LiAlHC12H27O3 weighed 12.7g. If Li atoms are conserved, what is the percentage yield.

1. The reaction is as follows:




1 mole of LiAlH_​4 reacts with 3 moles of tert-butyl alcohol.


So, Moles of tert-butyl alcohol reacted = 0.05 (3) = 0.15 moles

 but we have 1 mole of tert-butyl alcohol,

  So % yield = $\frac{0.15}{1}\times 100 = 15 \%$