How many grams of O2 are required to burn 5.15 gal. of C8H18?

Question

The complete combustion of octane, C8H18, a component of gasoline, proceeds as is shown below.

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l)

(a) Octane has a density of 0.692 g/mL at 20°C. How many grams of O2 are required to burn 5.15 gal. of C8H18?

Answer 1

it looks like the hard part is already done (somebody balanced the equation and looked up the density for you, sweet)

5.15 gal*(3785 ml/gal)*(.692 g/ml)*(mole/114 g) = 118 moles octane

note that i had to look up the conversion for gallons to ml and I had to calculate the molecular weight of C8H18, I did it in my head with round atom weights, you may want to go back and do it better

okay, you have the moles of octane involved and the equation tells us that we need 25 moles of O2 for every 2 moles of octane

that would be 118*25/2=1479 moles of O2

using 32 g/mol as the molecular weight of O2

1479 moles O2*(32 g/mol)=47330 g O2

a quick reality check to see if we believe 47+ thousand grams
well the 5+ gallons of octane would weight about 40 lbs
47330/454=app 100 lbs

that seems high but believeable
I'm going with 47330 gms

you ought to go over this and check my work
there could easily be an arithmetic error in it
but this should give you a feel for how to do this kind of problem

good luck

Answer 2

I am bad at chemistry, but good with search engines... Here is a nice web site with tips on balancing Equations for a Chemical change.

http://antoine.frostburg.edu/chem/senese/101/reactions/balancing.shtml

And I also searched and found that for
100 grams of Octance the amount of O2 required is 350.88g and the balanced equation for the same will be
2 C8H18 + 25 O2 ---------> 16 CO2 + 18 H2O (as you have already mentioned)

The atomic weights involved in this problem are:
H: 1.0 g
C: 12.0 g
O: 16.0 g

Njoy....

Answer 3

evidently like the complicated area is already executed (someone balanced the equation and appeared up the density for you, sweet) 5.15 gal*(3785 ml/gal)*(.692 g/ml)*(mole/114 g) = 118 moles octane observe that i had to seem up the conversion for gallons to ml and that i had to calculate the molecular weight of C8H18, I did it in my head with round atom weights, you need to prefer to pass again and do it more effective ok, you've the moles of octane in touch and the equation tells us that we choose 25 moles of O2 for each 2 moles of octane that ought to nicely be 118*25/2=1479 moles of O2 making use of 32 g/mol because the molecular weight of O2 1479 moles O2*(32 g/mol)=47330 g O2 a short reality examine to be certain if we trust 40 seven+ thousand grams nicely the 5+ gallons of octane ought to weight about 40 lbs 47330/454=app one hundred lbs that looks extreme yet believeable i pick 47330 gms you need to prefer to omit this and examine my artwork there ought to extremely be an mathematics blunders in it yet this may furnish you with a experience for a thanks to attempt one of those challenge good success

Answer 4

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