Questions: The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2: C6H12O6 → 2 C2H5OH + 2 CO2 a. How many moles of CO2 are produced when 0.65 mol of C6H12O6 reacts in this fashion? b. How many grams of C6H12O6 are needed to form 8.49 g C2H5OH? c. How many grams of CO2 form when 3.25 g of C2H5OH are produced?

Solution Steps

The balanced chemical equation for the fermentation of glucose is: \[ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} + 2 \mathrm{CO}_{2} \]

Given: 0.65 mol of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$

From the balanced equation, 1 mole of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ produces 2 moles of $\mathrm{CO}_{2}$.

Therefore, 0.65 mol of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ will produce: \[ 0.65 \, \text{mol} \times 2 = 1.30 \, \text{mol of } \mathrm{CO}_{2} \]

\[ \boxed{1.30 \, \text{mol of } \mathrm{CO}_{2}} \]

Given: 8.49 g of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

First, calculate the molar mass of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$: \[ \text{Molar mass of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = 2(12.01) + 6(1.008) + 16.00 = 46.068 \, \text{g/mol} \]

Calculate moles of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$: \[ \text{Moles of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = \frac{8.49 \, \text{g}}{46.068 \, \text{g/mol}} = 0.1843 \, \text{mol} \]

From the balanced equation, 2 moles of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ are produced from 1 mole of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$.

Therefore, moles of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ needed: \[ \text{Moles of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = \frac{0.1843 \, \text{mol}}{2} = 0.09215 \, \text{mol} \]

Calculate the mass of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$: \[ \text{Molar mass of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = 6(12.01) + 12(1.008) + 6(16.00) = 180.156 \, \text{g/mol} \]

\[ \text{Mass of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = 0.09215 \, \text{mol} \times 180.156 \, \text{g/mol} = 16.60 \, \text{g} \]

\[ \boxed{16.60 \, \text{g of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}} \]

Given: 3.25 g of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

First, calculate moles of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$: \[ \text{Moles of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = \frac{3.25 \, \text{g}}{46.068 \, \text{g/mol}} = 0.07056 \, \text{mol} \]

From the balanced equation, 2 moles of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ produce 2 moles of $\mathrm{CO}_{2}$.

Therefore, moles of $\mathrm{CO}_{2}$ produced: \[ \text{Moles of } \mathrm{CO}_{2} = 0.07056 \, \text{mol} \]

Calculate the mass of $\mathrm{CO}_{2}$: \[ \text{Molar mass of } \mathrm{CO}_{2} = 12.01 + 2(16.00) = 44.01 \, \text{g/mol} \]

\[ \text{Mass of } \mathrm{CO}_{2} = 0.07056 \, \text{mol} \times 44.01 \, \text{g/mol} = 3.105 \, \text{g} \]

\[ \boxed{3.105 \, \text{g of } \mathrm{CO}_{2}} \]

Final Answer