Question

current attempt in progress
the complete combustion of ethanol, ( c_{2}h_{5}oh ), in fuels such as those added to gasoline could be represented as:
( 2c_{2}h_{5}oh + 7o_{2}
ightarrow 4co_{2}+6h_{2}o )
75.0 g of ethanol reacted with 105 g of oxygen in an experiment. 65.0 g of ( co_{2} ) was obtained. what is the percent yield for this reaction?
61.9%
78.8%
86.7%
90.6%
45.3%

Explanation:

Step1: Calculate moles of reactants

• Molar mass of \(C_2H_5OH\): \(M = 2\times12 + 6\times1+16=46\space g/mol\). Moles of \(C_2H_5OH\), \(n_{C_2H_5OH}=\frac{75.0\space g}{46\space g/mol}\approx1.63\space mol\)
• Molar mass of \(O_2\): \(M = 32\space g/mol\). Moles of \(O_2\), \(n_{O_2}=\frac{105\space g}{32\space g/mol}\approx3.28\space mol\)

Step2: Determine limiting reactant

From the balanced equation \(2C_2H_5OH + 7O_2
ightarrow4CO_2+6H_2O\), the mole ratio of \(C_2H_5OH\) to \(O_2\) is \(2:7\).
If \(n_{C_2H_5OH} = 1.63\space mol\), moles of \(O_2\) required \(n_{O_2}^{req}=1.63\times\frac{7}{2}=5.71\space mol\). Since \(3.28\space mol<5.71\space mol\), \(O_2\) is the limiting reactant.

Step3: Calculate moles of \(CO_2\) produced (theoretical)

Using the mole ratio of \(O_2\) to \(CO_2\) (\(7:4\)). Moles of \(CO_2\) (theoretical), \(n_{CO_2}^{theo}=3.28\times\frac{4}{7}\approx1.87\space mol\)

Step4: Calculate mass of \(CO_2\) (theoretical)

Molar mass of \(CO_2\): \(M = 44\space g/mol\). Mass of \(CO_2\) (theoretical), \(m_{CO_2}^{theo}=1.87\times44 = 82.3\space g\)

Step5: Calculate percent - yield

Percent - yield formula: \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\)
\(\text{Percent Yield}=\frac{65.0\space g}{82.3\space g}\times100\%\approx78.8\%\)

Answer:

78.8%