Question

The octane in gasoline burns according to the following equation:2C8H18 + 25O2 16CO2 + 18H2O(a) How many moles of CO2 can form from 0.603 mol of octane?(b) How many moles of water are produced by the combustion of 8.72 mol of octane?(c) If this reaction is used to synthesize 8.90 mol of CO2, how many moles of oxygen are needed?(d) How many moles of octane?

Answer 1

The balanced equation of the octane gasoline is as follows :

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

(a) Calculate moles of CO2 that can form from 0.603 mol of octane:

• From the balanced equation, we can see that :

2 moles of octane : 16 moles of CO2

then, 0.603mol octane : x moles of CO2....( do a cross multiply )

∴ xmoles CO2 = (0.603 mol octane * 16mole CO2) / 2mole Octane

= 4.824 moles

This means that 4.824 moles of CO2 is formed from 0.603 mol octane

(b) Calculate moles of water are produced by the combustion of 8.72 mol of octane.

• 2 moles of octane gives 18 moles H2O

then, 8.72 moles Octane will give x moles of H2O

∴Moles of water = (18 * 8.72 )/2

= 78.48 moles

This means that 78.48 moles of H2O is formed from 8.72 mol octane

(c) Calculate moles of oxygen needed to synthesize 8.90 mol of CO2

• 16 moles CO2 require 25 moles O2

then, 8.90 mol CO2 requires x moles of O2

∴Moles of O2 = (8.90 *25 )/16

=13.91 moles

This means that 13.91 moles of O2 is needed to synthesize 8.90 mol CO2.

(d) Calculate moles of octane?

• 16 moles of CO2 require 2 moles octane

then , 8.90 CO2 require x moles of Octane

∴ Moles of octane = 8.90 * 2/16

=1.11 moles of Octane