Question bank/JEE Main/Chemistry

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).

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$Li{AlH}_{4} + t - butyl\ alcohol\overset{Ether}{\longrightarrow}{LiAlHC}_{12}H_{27}O_{3}(M \cdot W. = 254)$
0.05 mole 12.7 gram
[IMAGE] Li atom remain conserved so
No. of mole of LiAlH4 = No. of mole of LiAlHC12H27O3
So No. of mole of LiAlHC12H27O3 = 0.05 yield [IMAGE] = 100

Source: Grass-G / CH-16 Chemistry Paper 25 Nov. Fisrt.docx