Question

The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were allowed to become saturated with water before used and that the 1.2 g magnesium was the limiting reagent, what percentage of the product is expected to be lost due to the water in the solvent if the ether is not anhydrous?

Answer 1

Step-by-step explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

(100ml - 1.468 ml)

= 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

Amount of water =
`(1.468 ml * 23 ml)/(98.532 ml)`

= 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

`Mg + H_(2)O \rightarrow Mg(OH)_(2)`

Molar mass of Mg = 24.305 g

Molar mass of
`H_(2)O` = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

Amount of Mg =
`(24.305 g * 0.3427 ml)/(18 g)`

= 0.462 g