Question

Combustion of Octane: 

C8H18 + O2 → CO2 + H2O 

 Question 1: What are the coefficients for the balanced reaction of the combustion of octane?

 Problem 2: If 20 g of octane combust with 20 g of oxygen, which is the limiting reagent?

Answer 1

Q1)

aC8H18 + bO2 --> cCO2 + dH2O

Assuming the stoichiometric ratios in the reaction to be a, b, c, and d, we have:
8a = c
18a = 2d
=> 9a = d
2b = 2c + d

Therefore,

2b = 2(8a) + 9a
=> 2b = 16a + 9a = 25a

=> b = 12.5 a

Therefore,
b = 12.5 a
c = 8a
d = 9a

The reaction is:

a C8H18 + 12.5a O2 --> 8a CO2 + 9a H2O

Multiplying by 2;
2a C8H18 + 25a O2 --> 16a CO2 + 18a H2O

or:
2 C8H18 + 25 O2 --> 16 CO2 + 18 H2O

Q2)

If there are 20g of Octane and 20g of Oxygen is present;

No. of moles of Octane = Mass/Mol. Mass = 20/114.232 = 0.175 mol
No of moles of Oxygen = Mass/Mol. Mass = 20/32 = 0.625 mol

Now, from the equation in Q1, for 2 mol of Octane, 25 mol of Oxygen is required for complete combustion. Since 20 grams of Octane is 0.175 mol, we would require 2.1875 mol of Oxygen for complete combustion. However, only 0.625 mol of Oxygen is present. Therefore, Oxygen is the limiting reagent.