Combustion stoichiometry problem?

Question

What mass of air, 20.47% oxygen by mass, would it take to completely combust 1.00 gallon of glycerin (density of glycerin = 1.2613 g/mL, formula = C3H8O2)

balanced equation:

C3H8O2 + 4O2 --------> 3CO2 + 4H2O

The mass of the air needs to be in kg.

The answer is 39.2 kg, I just don't know how to get to there. I figured out the mass of the glycerin but I don't know where to go from there. Please help?

Answer 1

1 US gallon = 3.78541178 litres

3.78541178 litres = 3785.41178mL.

so you have  3785.41178mL glycerin combusting but we need it in moles.. to get there we use the density to figure out the mass and then use the mass and the molecular mass of glycerin to get the number of moles.  like this

3785.41178mL* 1.2613g/mL = 4774.539878g of glycerin

That formula is not correct C3H5(OH)3 is correct formula note there is one more oxygen (ref wikipedia)

the molecular weight is 92.09382 g/mol

therefore

4774.539878g / (92.09382g/mol) = 51.84430mol

ratio of glycerin to oxymen 1 1:4 therfore



moles of O2 in rxn is: 51.84430mol * 4= 207.37721mol O2

mass of oxygen:  32.00g/mol * 207.37721mol  =6636.0707g of oxygen

but they arent asking mass of oxygen they are asking mass of air.  

(6636g/x) = (20.47/100)  where x is mass of air

663600g = 20.47x

x= 32418g

which of course is 32.4 kg.

hmmm.. diff from your given answer

do it exactly the same but use molecular weight of glycerin of 76.0958g/mol instead of 92.09382 g/mol.

That will give you the "right" answer according to your folrmula that is missing an oxygen.

Cheers