Question

question 8
ether, (c₂h₅)₂o, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.
2 c₂h₅oh + h₂so₄ → (c₂h₅)₂o + h₂so₄ + h₂o
if 19.1 g c₂h₅oh reacts with excess h₂so₄, what is the theoretical yield of (c₂h₅)₂o?
g (c₂h₅)₂o
if this reaction actually produces 9.56 g (c₂h₅)₂o, what is the percent yield of (c₂h₅)₂o?
%

Explanation:

Step1: Calculate moles of \(C_2H_5OH\)

Molar mass of \(C_2H_5OH\) (\(M\)): \(M=(2\times12.01)+(6\times1.01)+16.00 = 46.08\space g/mol\)
Moles of \(C_2H_5OH\) (\(n\)): \(n=\frac{m}{M}=\frac{19.1\space g}{46.08\space g/mol}\approx0.4145\space mol\)

Step2: Determine moles of \((C_2H_5)_2O\)

From the balanced equation \(2C_2H_5OH
ightarrow(C_2H_5)_2O\), mole ratio \(n((C_2H_5)_2O):n(C_2H_5OH)=1:2\)
Moles of \((C_2H_5)_2O\): \(n((C_2H_5)_2O)=\frac{0.4145\space mol}{2}=0.20725\space mol\)

Step3: Calculate theoretical yield of \((C_2H_5)_2O\)

Molar mass of \((C_2H_5)_2O\) (\(M'\)): \(M'=(4\times12.01)+(10\times1.01)+16.00 = 74.14\space g/mol\)
Theoretical yield (\(m_{theo}\)): \(m_{theo}=n\times M'=0.20725\space mol\times74.14\space g/mol\approx15.37\space g\)

Step4: Calculate percent yield

Percent yield formula: \(\text{Percent Yield}=\frac{m_{actual}}{m_{theo}}\times100\%\)
Given \(m_{actual} = 9.56\space g\) and \(m_{theo}\approx15.37\space g\)
\(\text{Percent Yield}=\frac{9.56\space g}{15.37\space g}\times100\%\approx62.2\%\)

Answer:

15.37
62.2