# After the reaction, how much octane is left?

• The chemical equation for the reaction is given below:

$C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O$

• In this reaction, $0.430$ moles of octane react with $0.890$ moles of oxygen.

The question aims to find the moles of octane left after the said reaction takes place. Octane is a hydrocarbon whose chemical formula is $C_8H_{18}$. In the given chemical reaction, combustion of octane takes place in which $1$ mole of octane reacts with $25/2$ moles of oxygen.

Mole is the standard unit of measurement used in chemistry, and it refers to a very large quantity of atoms, molecules, or other particles. The mole of a substance corresponds to the $6.0223 x 10^{23}$ particles of that substance.

The question has specified the number of moles of oxygen and octane consumed during the reaction. The amount of octane left is to be determined.

The chemical equation is given below:

$C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O$

This balanced chemical equation can be simplified by multiplying with $2$ on both sides. The simplified chemical equation for this reaction is given below:

$2C_8H_{18} + 25 O_2 \rightarrow 16CO_2 + 18H_2O$

The chemical equation states that $2$ moles of octane react with $25$ moles of oxygen to produce $16$ moles of carbon dioxide and $18$ moles of water.

According to the question statement, $0.890$ moles of oxygen are consumed for a complete reaction. Therefore, the moles of octane consumed are:

$\text{2 moles of C_8H_{18}} \leftrightarrow \text{25 moles of O_2}$

$\text{moles of C_8H_{18}} \leftrightarrow \text{0.890 moles of O_2}$

$\text{moles of C_8H_{18}} = (2 x 0.890)/25$

$\text{moles of C_8H_{18} consumed} = 0.0712 mol$

As $0.430$ moles of octane are given and only $0.0712$ moles are consumed in the reaction, then the moles of octane left are:

$\text{moles left} = 0.430 mol – 0.0712 mol$

$\text{moles of C_8H_{18} left} = 0.3588 mol$

## Alternative Solution

The given balanced chemical equation is:

$C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O$

Where $1$ mole of octane reacts with $25/2$ moles of oxygen.

The moles of oxygen consumed in the reaction is $0.890$ mol. So the moles of octane consumed are:

$\text{moles of C_8H_{18} consumed} = 0.890/25/2 = 0.0712 mol$

$\text{moles of octane left} = 0.430 mol – 0.0712 mol$

$= 0.3588 mol$

## Example

The chemical equation for the reaction is given below:

$C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O$

In this reaction, $0.430$ moles of octane react with $0.890$ moles of oxygen. Calculate the moles of water produced.

Here, the moles of water produced need to be calculated.

For $25/2$ moles of oxygen, $9$ moles of water are produced.

$\text{9 moles of H_2O} \leftrightarrow \text{25/2 moles of O_2}$

For $0.890$ moles of oxygen, the moles of water produced is:

$\text{moles of H_2O} \leftrightarrow \text{0.890 moles of O_2}$

$\text{moles of H_2O} = (2 x 0.890 x 9)/25$

$\text{moles of H_2O produced} = 0.6408 mol$