Ethanol can be produced commercially by the hydration of ethylene. C_2H_4 + H_2O

Question

Ethanol can be produced commercially by the hydration of ethylene. C_2H_4 + H_2O rightarrow C_2H_5OH Some of the product is converted to diethyl ether in the aide reaction: 2C_2H_54OH rightarrow (C_2H_5)_2O + H_2O The feed to the reactor contains ethylene, steam and an inert gas. A sample of the reactor effluent gas is analyzed and found to contain 43.3 mol% ethylene, 2.5 mol% ethanol, 0.14 mol% ether, and 9.3 mol% inerts. Determine the molar now rates of each component in both the feed and effluent streams required to for the effluent stream to contain 100 kg/min of ethanol (see data in problem (:.) below). Liquid ethanol (see data in problem (:.) below) is fed to a space heater at a rate of 12.0 L/hr and burned with excess air. The product gas is analyzed and the following dry-basis mole percentages are determined: C_2H_5OH = 0.45%. CO_2 = 9.03%, and CO = 1.81%. Determine the molar flow rates of each component in both the feed and effluent streams. Maltose is converted (via fermentation) to ethanol by the following reaction: C_2H_22O_11 + H_2O(+yeast) rightarrow 4C_2H_5OH + 4CO_2(+yeast + H_2O) For every 1 kg of ethanol formed, 0.0794 kg of yeast is produced in the above reaction. For every 1 kg of yeast formed, 0.291 kg of water is produced in the above reaction. 80-proof whiskey produced by this process contains 40% (by volume) ethanol. Com fed to the process contains 60.84 wt% starch (precursor of maltose) and 15.5 wt% water. It is estimated that 101.2 bushels of com can be harvested from an acre of com, that each bushel is equivalent to 25.4 lb_m of com, and that 6.7 kg of ethanol can be obtained from a bushel of com. Assume whiskey contains primary only ethanol and water. The specific gravity of ethanol is 0.789. The molecular weights of water and ethanol are 18.016 g/mol and 46.07 g/mol respectively. Assume a fractional conversion of maltose of 0.89. What acreage of farmland is required to produce 10^5 barrels (1 barrel = 119.240 L) of 80-proof whiskey per year? What is the mole fraction of ethanol in 80-proof whiskey? How much yeast would be formed (in kg) in producing 10^6 barrels of 80-proof whiskey? How many moles of CO_2 would be produced by making 10^6 barrels of 80-proof whiskey? How many moles of maltose would be required to make 10^5 barrels of 80-proof whiskey?

Explanation / Answer

a)

The 80-proof whiskey contains 40 vol% ethanol and 60 vol% water.

106 barrel is 119.240 *106 L of 80-whiskey, out of which 40 vol% is ethanol. i.e. 47.696*106 L of ethanol.

the specific gravity of ethanol is 0.789 hence the weight of ethanol is (47.696*106*0.789=)37.632*106 Kg

1 bushel is equivalent to 25.4lb corn is 6.7kg ethanol

hence 37.632106 kg of ethanol is equivalent to 5.616*106 bushels

1 acre of land harvests 101.2 bushels

Hence 5.616 bushels are harvested in (5.616*106/101.2=) 55501.15 acre.

Hence farmland required to produce 106 barrels of 80-proof whiskey is 55501.15 acre.

b)

80-proof whiskey contains 40 vol% ethanol and 60 vol% water

let's assume 100 Litre of 80-proof whiskey which contains 40L ethanol and 60L water

specific gravity of ethanol is 0.789 and water is 1

weight of ethanol is 0.789*40 =31.56 kg

weight of water is 60*1 = 60 kg

Molecular wight of ethanol is 46.07 kg/kmol and water is 18.016kg/kmol

moles of ethanol is 31.56/46.07 = 0.685 kmol

moles of water is 60/18.06 = 3.322 kmol

mole fraction of ethanol is (0.685 / (0.685 + 3.322)) = 0.171

mole fraction of ethanol in 80-proof whiskey is 0.171

c)

The 80-proof whiskey contains 40 vol% ethanol and 60 vol% water.

106 barrel is 119.240 *106 L of 80-whiskey, out of which 40 vol% is ethanol. i.e. 47.696*106 L of ethanol.

the specific gravity of ethanol is 0.789 hence the weight of ethanol is (47.696*106*0.789=)37.632*106 Kg

For 1 kg of ethanol formed, 0.0794 kg of yeast is produced.

Hence for 37.632*106 Kg ethanol, yeast produced is 2.988*106 Kg

Hence yeast formed in producing 106 barrel of 80-proof whiskey is 2.988*106 Kg   

d)

80-proof whiskey contains 40 vol% ethanol and 60 vol% water

106 barrel i.e. 119.240*106 Litre of 80-proof whiskey which contains 47.696*106L ethanol.

specific gravity of ethanol is 0.789

weight of ethanol is 0.789*47.696*106L =56.448*106 kg

Molecular wight of ethanol is 46.07 kg/kmol

moles of ethanol is 56.448*106/46.07 = 1.225*106 kmol

In the fermaentation reaction provided, according to stoichometry, every mole of maltose ferments to produce 4 moles of ethanol and CO2 each, i.e. equivalent moles of ethanol and CO2 are formed during fermentation.

hence 1.225*106 kmol of CO2 would be produced by making 106 barrels of 80-proof whiskey

e)

80-proof whiskey contains 40 vol% ethanol and 60 vol% water

106 barrel i.e. 119.240*106 Litre of 80-proof whiskey which contains 47.696*106L ethanol.

specific gravity of ethanol is 0.789

weight of ethanol is 0.789*47.696*106L =56.448*106 kg

Molecular wight of ethanol is 46.07 kg/kmol

moles of ethanol is 56.448*106/46.07 = 1.225*106 kmol

In the fermaentation reaction provided, according to stoichometry, every mole of maltose ferments to produce 4 moles of ethanol and CO2 each, i.e. 1 mole of maltose fermented is equivalent of 4 moles of ethanol produced.

1.225*106 kmol of ethanol is produced

hence maltose required is (1.225*106 /4 =)0.306*106 kmol

hence 0.306*106 kmol of maltose would be produced by making 106 barrels of 80-proof whiskey.

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