Answer to Question #15621 in Inorganic Chemistry for Meghan

Question

In the 2011-2012 academic year, Dr. Harmon used approximately 515 kg of methane in his home for heating, hot water, and stove/oven use. How many 75-gallon barrels of waste corn oil should Dr. Harmon collect to produce his own methane for 1 entire year assuming he uses the exact same amount of methane as in the 2011-2012 academic year? The density of oleic and linoleic acids are very similar, you may assume both are 0.90 g/cm3. Additionally, you may assume the percent yield of methane production from corn oil pyrolysis is 100%

HC18H31O2+ O2(g) àCH4(g) + CO2(g) + H2O(g) + H2(g)

HC18H33O2 + O2(g) àCH4(g) + CO2(g) + H2O(g) + H2(g)

To make this problem easier, you may assume 65% of the methane produced comes from linoleic acid and 35% of the methane produced comes from oleic acid.

Answer 1

HC₁₈H₃₁O₂(l) + 12.5O₂(g)  6CH₄(g) + 12CO₂(g) + 3H₂O(g) + H₂(g) 1 mole of acid --- 6 moles of Methane.

m of CH₄ is 580 * 65% / 100% = 377 kg
n of CH₄ = 377000/16 = 23562 mol
n of acid = 23562/6 = 3927 mol
3927 * 280 = 1099560 g
1099560 / 0.9 = 1221733 cm³ = 1221.73 L
1221L/3.79 L/gal = 322.35 gallons

1 barrel = 75 gallons
x barrels = 322.35 gallons

x = 4.298 = 4.3 barrels.

HC₁₈H₃₃O₂(l) + 11O₂(g) ==> 6CH₄(g) + 12CO₂(g) + 4H₂O(g) + H₂(g)


1 mole of acid --- 6 moles of Methane.



m of CH₄ is 580 * 35% / 100% = 203 kg
n of CH₄ = 203000/16 = 12687.5 mol
n of acid = 2114.58 mol
v = 662.569 L = 174.8 gallons

1 barrel = 75 gallons

x barrels = 174.8 gallons



x = 2.33 barrels.