Question

0.05 mole of LiAlH4 in ether solution was placed in flask containing 74g of t-butyl alcoho.the product LiAlHC12H27O3 weighed 12.7g .if Li atoms are conserved,the percentage yield is ​

Answer 1

Answer:

The percentage yield is ​100%.

Explanation:

$LiAlH_4+3(CH_3)_3-C-OH\rightarrow LiAlHC_{12}H_{27}O_3+3H_2$

Moles of t-butyl alcohol :

$\frac{74 g}{74 g/mol}= 1mol$

According to reaction, 1 mole of lithium aluminum hydride gives 1 mole of $LiAlHC_{12}H_{27}O_3$

Then 0.05 mol of lithium aluminum hydride will give

$\frac{1}{1}\times 0.05 =0.05 mol$

Theoretical yield of $LiAlHC_{12}H_{27}O_3$:

$0.05 mol \times 254 g/mol=12.7 g$

Experimental yield = 12.7 g

The percentage yield :

$=\frac{Experimental}{Theoretical}=\frac{12.7 g}{12.7 g}\times 100 =100\%$

Answer 2

Answer:

Explanation:Applying POAC on Li

1 × moles of LiAlH4 = 1× moles of LiAlH C12H27O3

254 × 0.32 = 1 × wt. of LiAlH C12H27O3.

wt. of LiAlH C12H27O3 = 81.28 g.