Question
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 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?
Step 1
We can do this by dividing the mass of glucose by its molar mass. The molar mass of glucose is $12 \times 6 + 12 + 16 \times 6 = 180 \, g/mol$. So, the number of moles of glucose is $\frac{5.97 \, g}{180 \, g/mol} = 0.033 \, mol$. Show more…
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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 reacts and $1.44 \mathrm{L}$ of $\mathrm{CO}_{2}$ gas is collected at 293 K and 0.984 atm, what is the percent yield of the reaction?
In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: if 5.97 g of glucose reacts and 1.44 l of co2 gas is collected at 293 k and 0.984 atm, what is the percent yield of the reaction?
In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide. C6H12O6 -----> 2C2H5OH + 2CO2 If 5.97 g of glucose are reacted and 1.44 L of CO2 gas are collected at 293 K and 0.984 atm, what is the percent yield of the reaction?
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