in alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: c6h12o6(s) → 2c2h5oh(l) +…

in alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: c6h12o6(s) → 2c2h5oh(l) + 2co2(g) if 7.06 g of glucose are reacted and 1.44 l of co2 gas are collected at 293.15 k and 0.984 atm, what is the percent yield of the reaction? be sure your answer has the correct number of significant digits. note: reference the fundamental constants table for additional information.

in alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: c6h12o6(s) → 2c2h5oh(l) + 2co2(g) if 7.06 g of glucose are reacted and 1.44 l of co2 gas are collected at 293.15 k and 0.984 atm, what is the percent yield of the reaction? be sure your answer has the correct number of significant digits. note: reference the fundamental constants table for additional information.

Answer

Explanation:

Step1: Calculate moles of glucose

The molar - mass of glucose ($C_6H_{12}O_6$) is $M=(6\times12.01 + 12\times1.01+6\times16.00)\ g/mol = 180.18\ g/mol$. The number of moles of glucose, $n_{glucose}=\frac{m}{M}=\frac{7.06\ g}{180.18\ g/mol}=0.0392\ mol$.

Step2: Determine theoretical moles of $CO_2$

From the balanced chemical equation $C_6H_{12}O_6(s)\rightarrow2C_2H_5OH(l)+2CO_2(g)$, the mole - ratio of glucose to $CO_2$ is 1:2. So the theoretical number of moles of $CO_2$, $n_{CO_2,theo}=2\times n_{glucose}=2\times0.0392\ mol = 0.0784\ mol$.

Step3: Calculate moles of collected $CO_2$ using the ideal gas law

The ideal gas law is $PV = nRT$, where $P = 0.984\ atm$, $V = 1.44\ L$, $T=293.15\ K$, and $R = 0.0821\ L\cdot atm/(mol\cdot K)$. Rearranging for $n$, we get $n=\frac{PV}{RT}$. Substituting the values: $n_{CO_2,collected}=\frac{0.984\ atm\times1.44\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times293.15\ K}=0.0587\ mol$.

Step4: Calculate percent yield

The percent - yield formula is $\text{Percent Yield}=\frac{n_{CO_2,collected}}{n_{CO_2,theo}}\times100%$. Substituting the values: $\text{Percent Yield}=\frac{0.0587\ mol}{0.0784\ mol}\times100% = 74.9%$.

Answer:

74.9%