$0.05$ mole of $L i A l H_{4}$ in ether solution was placed in a flask containing $74 g$ (1 mole) of t - butyl alcohol. The product $L i A l H C_{12} H_{27} O_{3}$ weighed $12.7 g$ .
If $L i$ atoms are conserved, the percentage yield is:
(Given atomic masses (amu): $L i = 7, A l = 27, H = 1, C = 12, O = 16$ )

25%

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75%

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100%

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15%

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Solution

The correct option is C 100%
$L i A l H_{4} + t - b u t y l a l c o h o l E t h e r - -- \to L i A l H C_{12} H_{27} O_{3} (M . W t . = 254)$

$0.05 m o l 12. 7 g$

so, moles of product $= \frac{12.7}{254} = 0.05 m o l$

$L i$ atom remains conserved so,
Number of moles of $L i A l H_{4}$ = Number of moles of $L i A l H C_{12} H_{27} O_{3}$
So, Number of moles of $L i A l H C_{12} H_{27} O_{3}$ that should form are $= 0.05$
% yield $= \frac{0.05}{0.05} \times 100 = 100$ %

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L i A l H_{4}

L i A l H C_{12} H_{27} O_{3}

The mass of 0.5 moles of Water molecules is:

(Atomic mass, H =1; O = 16)

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H = 1, N = 14, O = 16

(N H_{4})_{2} C O_{3}

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C O_{2}

N a

N a C O_{3}

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C = 12, O = 16, H = 1, N a = 23, A l = 27, N = 14

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