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# Answer to Question #83751 in Inorganic Chemistry for Joe

Question #83751
I have a can of diethylene glycol that will burn complete in 6 hours. How much air (with 21% O2) will be required to burn all of the 7.41 fl. ounces of fuel?
1
2018-12-14T05:56:06-0500

Answer on Question #83751, Chemistry/ Inorganic Chemistry

I have a can of diethylene glycol that will burn complete in 6 hours. How much air (with 21% O2) will be required to burn all of the 7.41 fl. ounces of fuel?

Solution

(HOCH2CH2)2O + 5O2 = 4CO2 + 5H2O

V((HOCH2CH2)2O) = 7.41 fl.ounces = 7.41/33.814 = 0.219 L

n= V/Vm

n((HOCH2CH2)2O) = 0.219 L/ 22.4 mol/L = 0.00978 moles

According to equation buring of 1 mole of diethylene glycol requires 5 moles of oxygen

We have 0.00978 moles of diethylene glycol which require x moles of oxygen

Solve the proportion:

1/0.00978 = 5/x

x= 0.0489

n(O2) = 0.0489 moles

V = n*Vm

V(O2) = 0.0489 moles *22.4 mol/L = 1.095 L

Ï†(ÎŸ2) = V(O2)/ Vair => Vair = V(O2)/Ï†(ÎŸ2) = 1.095 L / 0.21 = 5.214 L

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