Question

Air is 23.19% o2 and 75.46% n2 by weight) would be needed to burn a pound of gasoline by a reaction whereby c8h18 reacts with o2 to form co2 and h2o?

Answer 1

Approximately
`15.10` pounds.

Start by balancing the equation for the complete combustion of octane in oxygen:

`\text{C}_8\text{H}_(18) + 25/2 \;\text{O}_2 \to 8 \; \text{CO}_2 + 9 \; \text{H}_2\text{O}`

`2 \; \text{C}_8\text{H}_(18) + 25 \;\text{O}_2 \to 16 \; \text{CO}_2 + 18 \; \text{H}_2\text{O}`

Thus it would take
`25` molecules of oxygen to completely react with
`2` molecules of octane. Oxygen and octane thus react at a ratio of

`25 \; \text{mol} \; \text{O}_2` to
`2 \; \text{mol} \; \text{C}_8 \text{H}_(18)`

Given the molar mass of the species:

• 
  `M(\text{O}_2) = 2 * 16.00 = 32.00 \; \text{g} \cdot \text{mol}^(-1)`
• 
  `M(\text{C}_8\text{H}_(18)) = 8 * 12.01 + 18 * 1.008 =  114.22 \; \text{g} \cdot \text{mol}^(-1)`

The mass ratio between the two species when the combustion proceeds to completion would thus equal:

`25 \; \text{mol} * 32.00 \; \text{g} \cdot \text{mol}^(-1) = 800.00 \; \text{g} \;` to

`2 \; \text{mol} * 114.22 \; \text{g} \cdot \text{mol}^(-1) = 228.45 \; \text{g}`

It would thus take

`\begin{array}{lll}m(\text{O}_2)& = & 1 \; \text{lb} \; \text{C}_8\text{H}_(18) * (800.00 \; \text{g}\; \text{O}_2) / ( 228.45 \; \text{g}\; \text{C}_8 \text{H}_(18))\\ &=& 3.5019 \; \text{lb} \; \text{O}_2 \end{array}`

to combust completely one pound of octane.

Apply the mass ratio stated in the question:

`\begin{array}{lll}m(\text{Air}) & = & m(\text{O}_2) / \frac{m(\text{O}_2)}{m(\text{Air})}\\ & = & 3.5019 \; \text{lb} / 0.2319 \\ & = & 15.10 \; \text{lb}\end{array}`