Questions: If 952 g C6H12O6 reacts according to the following equation, how many liters of C2H5OH can be produced, assuming 100% yield? Density of C2H5OH is 0.789 g / mL C6H12O6 is 180.2 g / mol. C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g) Hint: you must convert C2H5OH to LC2H5OH as last step. Hint: 1000 mL=1 L 7.32 L 551 L 0.0925 L 0.217 L 93.0L 0.617 L 4.04 L 1.59 L

Solution Steps

First, we need to calculate the number of moles of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ using its given mass and molar mass.

\[ \text{Moles of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = \frac{952 \, \text{g}}{180.2 \, \text{g/mol}} = 5.2841 \, \text{mol} \]

According to the balanced chemical equation, 1 mole of $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ produces 2 moles of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$. Therefore, the moles of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ produced are:

\[ \text{Moles of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = 2 \times 5.2841 \, \text{mol} = 10.5682 \, \text{mol} \]

The molar mass of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ is approximately 46.08 g/mol. Thus, the mass of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ is:

\[ \text{Mass of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = 10.5682 \, \text{mol} \times 46.08 \, \text{g/mol} = 486.9 \, \text{g} \]

Using the density of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, which is 0.789 g/mL, we convert the mass to volume in liters:

\[ \text{Volume in mL} = \frac{486.9 \, \text{g}}{0.789 \, \text{g/mL}} = 617.0 \, \text{mL} \]

Convert mL to L:

\[ \text{Volume in L} = \frac{617.0 \, \text{mL}}{1000} = 0.617 \, \text{L} \]

Final Answer