Consider this reaction: C8H18 + O2 -> CO2 + H2O?

Question

Under standard conditions, the volume of air in liters required for complete combustion of 228 grams of octane is about?

A. 22.4
B. 2.50
C. 560
D. 1560
E. 2800

And why please because I don't understand how to solve this.

Answer 1

First of all: What is the molecular weight of octane?
It is 8*12 + 18 = 114. SO 2 moles of octane are being burnt.
Now: How much oxygen is needed to combust 2 moles of octane.

(This is the equation balancing the C,H,Os)
2C8H18 + 25O2 ---> 16CO2 + 18H2O


We need 25 moles of Oxygen
How much volume does a mole of a gas occupy at STP? From the question I assume it is 22.4 (long time since I did chemistry)
AND What percentage of the air is Oxygen? 20%?
If so the answer is 25 * 22.4 * 5 i.e. E

Answer 2

1.00 g octane x (1 mol / 114 g) = 0.00877 mol octane The reaction is: 2C8H18 + 25O2 ---> 16CO2 + 18H2O 2 mol octane reacts with 25 mol O2. mol O2 = 0.00877 mol octane x (25 mol O2 / 2 mol octane) = 0.110 mol O2 Now use the ideal gas law to find the volume. PV = nRT (use units consistent with those on R) V = nRT/P 98.7 kPa x (1.00 atm / 101.3 kPa) = 0.974 atm V = (0.110 mol x 0.08206 L atm / K mol x 297 K) / 0.974 atm V = 2.75 L

Answer 3

1. balance the equation
2. work out the moles of octane
3. appreciate the mole ratio of 1:12.5
4. go for 25 moles of oxygen!

Now do the rest.