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Chapter 5: Problem 56

In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) $$ If \(5.97 \mathrm{~g}\) of glucose are reacted and \(1.44 \mathrm{~L}\) of \(\mathrm{CO}_{2}\) gas are collected at \(293 \mathrm{~K}\) and \(0.984 \mathrm{~atm},\) what is the percent yield of the reaction?

Short Answer

Expert verified
The percent yield of the reaction is 90.94%.

Step by step solution

01

Determine theoretical yield

The balanced chemical reaction indicates that one molecule of glucose (\(C_{6}H_{12}O_{6}\)) reacts to form two molecules of carbon dioxide (\(CO_{2}\)). Therefore, the molar ratio of glucose to carbon dioxide is 1:2. This means that the mass of glucose (5.97g) needs to be converted into moles using glucose's molar mass (180.156g/mol): \[\frac{5.97g}{180.156g/mol} = 0.0331mol\] The theoretical number of moles of carbon dioxide that should be formed from this reaction is therefore: \(0.0331mol \times 2 = 0.0662mol\)
02

Calculate actual yield

Using the ideal gas law, the moles of carbon dioxide can be converted from the given volume. Given: pressure, \(P = 0.984 atm\); temperature, \(T = 293K\); volume, \(V = 1.44L\) and R (ideal gas constant) = 0.0821 L·atm/K·mol. The ideal gas law is: \(PV = nRT\) Solving for n (moles of gas): \[n = \frac{PV}{RT} = \frac{(0.984 atm) (1.44 L)}{(0.0821 L·atm/K·mol) (293 K)} = 0.0602mol\]
03

Determine percent yield

Percent yield of a reaction can be calculated using the following formula: \[(percent yield) = \frac{(actual yield)}{(theoretical yield)} \times 100\% \rightarrow (percent yield) = \frac{0.0602mol}{0.0662mol} \times 100\% = 90.94\%\]

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