Answer:
87.64% is the percent yield of ether.
Explanation:
[tex]C_2H_5OH + H_2SO_4\rightarrow (C_2H_5)_2O + H_2SO_4.H_2O[/tex]
Volume of an ether,v= 1.17 L = 1170 mL
Experimental mass of an ether = m
Density of an ether ,d= 0.7134 g/mL
[tex]m=d\times v=0.7134 g/mL\times 1170 mL=834.6780 g[/tex]
Experimental yield of an ether = 834.678 g
Volume of an alcohol = V = 1.500 L = 1500 mL
Mass of an alcohol = M
Density of an alcohol ,D= 0.7894 g/mL
[tex]M=D\times V=0.7894 g/mL\times 1500 mL=1,184.1 g[/tex]
Moles of an alcohol = [tex]\frac{1,184.1 g}{46 g/mol}=25.74 mol[/tex]
According to recatiopn 2 moles of an alcohol gives 1 mole of an ether.
Then 25.74 moles of alcohol will give:
[tex]\frac{1}{2}\times 25.74 mol=12.87 mol[/tex] of an ether.
Mass of 12.87 moles of an ether = 12.87 mol =74 g/mol = 952.38 g
Theoretical yield of the an ether = 952.38 g
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]\%\text{ yield}=\frac{834.678 g}{952.38 g}\times 100=87.64\%[/tex]
87.64% is the percent yield of ether.
The percent yield of this reaction is 87%.
We define the percent yield as the ratio of the actual yield to the theoretical yield multiplied by 100.
Mass of ether = 0.7134 g/mL * 1170 mL = 835 g
Mass of C2H5OH = 0.7894 g/mL * 1500 mL = 1184 g
Number of moles of C2H5OH = 1184 g/46 g/mol =26 moles
Since the reaction is 2:1, number of moles of ether produced = 13moles
Theoretical yield of ether = 13 moles * 74 g/mol = 962 g
Percent yield = 835 g/ 962 g * 100/1 = 87%
Learn more about percent yield: https://brainly.com/question/916827
