Question

0.05 mole of $LiAl{H}_4$ in ether solution was placed in a flask containing 74g (1 mole ) of t-butyl alcohol. The product $LiAlH{C}_{12}{H}_{27}{O}_3$ weighed 12.7 g. If Li atoms are conserved, the percentage yield is : (Li = 7, Al = 27, H = 1, C = 12, O = 16).

• A. 25%
• B. 75%
• C. 100 %
• D. 15 %

Answer 1

The correct option is C 100 %

Reaction:
$LiAl{H}_4+3(C{H}_3{)}_{3}C-OH\to LiAlH{C}_{12}{H}_{27}{O}_3+3H_2$
1 mol 3 mol 1 mol
0.05 mol 0.15 mol 0.05 mol
Now,

Molar mass of t-butyl alcohol $=74g/mol$

Number of moles of t-butyl alcohol $=\frac{74g}{74g/mol}=1mol$

$\therefore$ $LiAl{H}_4$ is the limiting reagent.
Again,
Moles of $LiAlH{C}_{12}{H}_{27}{O}_3$ obtained theoretically $=0.05mol$
Mass of $LiAlH{C}_{12}{H}_{27}{O}_3$ obtained theoretically $=$ moles $\times$ Molar mass $=0.05\times 254=12.7g$
Mass of $LiAlH{C}_{12}{H}_{27}{O}_3$ obtained experimentally $=12.7g$

$\therefore \%$ yield $=\frac{\text{experimentally mass}}{\text{theoretically mass}}\times 100=\frac{12.7}{12.7}\times 100=100\%$