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

Question

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

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?

C8H18 is 114.232 grams per mole
O2 is 32 grams per mole
CO2 is 44.011 grams per mole
H2O is 18.02 grams per mole

Answer 1

I don't know the exact conversion of gallon to liter that you want to use . Considering that 1 US gallon =3.785411784 liters

d=m/V => m=d*V= 0.692 * 5.15 * 3.785411784 *10^3 =>
m=13490 g

From the stoichiometry

2*114.232 g octane require 25*32 g O2
13490 g octane require x g O2

thus x= 25*32*13490/(2*114.232) =47237 g

Answer 2

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 3

You did not convert it to moles, it really is what you want it to be in for stoichiometry. you also choose the balanced stoichiometry to draw close theoretical ratios. 2C8H18 + 25O2 ---> 16CO2 + 18H2O Now you comprehend you want 25 moles O2 for 2 moles C818. on account that C8H18 has a molar mass of 114.23g, divide that into 3143.4g to get moles. 3143.4g (1mol/114.23g) = 27.fifty 2 moles C8H18 you are able to now play with the ratio: 2:25 is a million:12.5. multiply via 27.fifty 2 --> 27.fifty 2 : 344 moles O2. Convert this to grams if it really is what that is meant to be in (18g/mol). good success!

Answer 4

What on Earth is this gallons business? Are from The Middle Ages?