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Chapter 3: Problem 72

Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+2 \mathrm{CO}_{2}$$ Starting with \(500.4 \mathrm{~g}\) of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process? (Density of ethanol \(=0.789 \mathrm{~g} / \mathrm{mL} .)\)

Short Answer

Expert verified
The maximum amount of ethanol that can be obtained from 500.4 g of glucose is 256.3 g of ethanol or 0.325 L of ethanol.

Step by step solution

01

Calculate the Molar Mass of Glucose and Ethanol

The molar mass of glucose (C6H12O6) and ethanol (C2H5OH) are calculated as follow: The molar mass of glucose is \(180.16 \mathrm{~g/mol}\) and the molar mass of ethanol is \(46.07 \mathrm{~g/mol}\)
02

Calculate the Amount of Glucose in Moles

The amount of glucose is given in grams and is converted into moles by dividing the given mass by the molar mass of glucose. This gives: \(\frac{500.4 \mathrm{~g}}{180.16 \mathrm{~g/mol}} = 2.78 \mathrm{~mol}\)
03

Calculate the Maximum Amount of Ethanol in Moles

From the stoichiometry of the reaction, 1 mol of glucose produces 2 mol of ethanol. Therefore, the maximum amount of ethanol produced is \(2.78 \mathrm{~mol} * 2 = 5.56 \mathrm{~mol}\)
04

Calculate the Maximum Amount of Ethanol in Grams

The maximum amount of ethanol in grams is given by the product of the number of moles and the molar mass of ethanol. This gives: \(5.56 \mathrm{~mol} * 46.07 \mathrm{~g/mol} = 256.3 \mathrm{~g}\)
05

Calculate the Maximum Amount of Ethanol in Liters

The maximum amount of ethanol in liters is determined by dividing the mass of ethanol by the density. This gives: \(\frac{256.3 \mathrm{~g}}{0.789 \mathrm{~g/mL}} = 325 \mathrm{~mL}\) or \(0.325 \mathrm{~L}\).

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Most popular questions from this chapter

Describe the steps involved in the mole method.

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