Question

for all complete reduction of 5.8g of acetone to isopropyl alcohol, the quantity of LiAlH4 required (assuming chemical yield to be 100%) is approximately​

Answer 1

Answer:

Reverse the reaction for Re hydration of pop to form gypsum

Then the reaction is

                                               (373%)

$Caso_{4} .2H_{2} O$  →        $Casox_{4} .\frac{1}{2} H_{2} O+1\frac{1}{2} H_{2} O$

Here gypsum will under goes re hydration then we get in the form of pop

Answer 2

Given:

Weight of acetone = 5.8 g

To find:

Quantity of LiAlH4 required to reduce acetone to isopropyl alcohol

Solution:

The reaction will be:

Acetone + LiAlH4 = isopropyl alcohol

From the reaction, we can conclude that 1 mole of acetone with 1 mole of LiAlH4 gives 1 mole of isopropyl alcohol.

Total moles of acetone = given weight of acetone/ molecular weight

= 5.8 / 58 = 1 / 10

Since the number of moles of LiAlH4 is equal to the moles of acetone, therefore, number of moles of LiAlH4 will be 1/10.

Since, moles = weight/ molecular weight

Molecular weight of LiAlH4 = 6.9 + 27+ (4*1)

1/10 = weight/ (6.9+27+4)

weight = 37.9 / 10

weight = 3.79 g

Therefore, the quantity of LiAlH4 required is 3.79 g.