Question

Diethyl ether is produced from ethanol according to the following equation: 2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l) Calculate the percent yield if 65.2 g of ethanol reacts to produce 17.2 g of ether.

Answer 1

Answer : The percent yield is, 32.79 %

Explanation :

First we have to calculate the moles of
`CH_3CH_2OH`.

`\text{Moles of }CH_3CH_2OH=\frac{\text{Mass of }CH_3CH_2OH}{\text{Molar mass of }CH_3CH_2OH}=(65.2g)/(46.07g/mole)=1.415mole`

Now we have to calculate the moles of
`CH_3CH_2OCH_2CH_3`

The balanced chemical reaction will be,

`2CH_3CH_2OH(l)\rightarrow CH_3CH_2OCH_2CH_3(l)+H_2O(l)`

From the balanced reaction, we conclude that

As, 2 moles of
`CH_3CH_2OH` react to give 1 mole of
`CH_3CH_2OCH_2CH_3`

So, 1.415 moles of
`CH_3CH_2OH` react to give
`(1.415)/(2)=0.7075` mole of
`CH_3CH_2OCH_2CH_3`

Now we have to calculate the mass of
`CH_3CH_2OCH_2CH_3`

`\text{Mass of ether}=\text{Moles of ether}* \text{Molar mass of ether}`

`\text{Mass of }ether=(0.7075mole)* (74.12g/mole)=52.44g`

The theoretical yield of ether,
`CH_3CH_2OCH_2CH_3` = 52.44 g

Now we have to calculate the percent yield of
`CH_3CH_2OCH_2CH_3`

`\%\text{ yield of ether}=\frac{\text{Actual yield of ether}}{\text{Theoretical yield of ether}}* 100=(17.2g)/(52.44g)* 100=32.79\%`

Therefore, the percent yield is, 32.79 %