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 $CH_3CH_2OH$.

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

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 $\frac{1.415}{2}=0.7075$ mole of $CH_3CH_2OCH_2CH_3$

Now we have to calculate the mass of $CH_3CH_2OCH_2CH_3$

$\text{Mass of ether}=\text{Moles of ether}\times \text{Molar mass of ether}$

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

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

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

$\%\text{ yield of ether}=\frac{\text{Actual yield of ether}}{\text{Theoretical yield of ether}}\times 100=\frac{17.2g}{52.44g}\times 100=32.79\%$

Therefore, the percent yield is, 32.79 %