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Answer: In alcohol fermentation, yeast converts glucose to
Chapter 5, Problem 56P(choose chapter or problem)
In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide:
\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+2 \mathrm{CO}_{2}(\mathrm{g})\)
If 5.97 g of glucose are reacted and 1.44 L of \(\mathrm{CO}_{2}\) gas are collected at 293 K and 0.984 atm, what is the percent yield of the reaction?
Questions & Answers
QUESTION:
In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide:
\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+2 \mathrm{CO}_{2}(\mathrm{g})\)
If 5.97 g of glucose are reacted and 1.44 L of \(\mathrm{CO}_{2}\) gas are collected at 293 K and 0.984 atm, what is the percent yield of the reaction?
ANSWER:Step 1 of 4
The goal of the problem is to calculate the percent yield of the reaction.
Given equation:
\(C_6H_{12}O_6 (s) \rightarrow 2C_2H_5OH (l) + 2CO_2 (g)\)
Given:
Mass of glucose = 5.97 g
V of \(CO_2\) = 1.44 L
T = 293 K
P = 0.984 atm