Answer to Question #83751 in Inorganic Chemistry for Joe

Question

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?

Answer 1

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/V_m

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*V_m

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

φ(Ο₂) = V(O₂)/ V_air => V_air = V(O₂)/φ(Ο₂) = 1.095 L / 0.21 = 5.214 L

Answer:V_air = 5.214 L