Question

Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide: C6H12O6 → 2C2H5OH + 2CO2 glucose ethanol Starting with 510.6 g of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process (density of ethanol = 0.789 g/mL)?

Answer 1

Answer:

Mass of ethanol obtained is 260.682 grams.

Volume of the ethanol obtained is 0.3304 L.

Explanation:

$C_6H_{12}O_6\rightarrow 2C_2H_5OH + 2CO_2$

Moles of glucose = $\frac{510.6 g}{180 g/mol}=2.833 mol$

According to the reaction, 1 mol of gluocse gives 2 moles of ethanol.

Then 2.833 moles of glucose will give :

$\frac{2}{1}\times 2.8333 mol=5.667 mol$

Mass of 5.667 moles of an ethanol :

5.667 mol × 46 g/mol = 260.682 g

Volume of ethanol = V

Density of the ethanol = 0.789 g/mL

$Density=\frac{Mass}{Volume}$

$0.789 g/mL=\frac{260.682 g}{V}$

$V=\frac{260.682 g}{0.789 g/mL}=330.40 mL = 0.3304 L$

Mass of ethanol obtained is 260.682 grams.

Volume of the ethanol obtained is 0.3304 L.