TEST YOURSELF 1.18

The combustion of octane [tex]\(\left( C_8H_{18} \right)\)[/tex] as fuel is represented by the equation below:

[tex]\[
2 C_8H_{18}(l) + 25 O_2(g) \rightarrow 16 CO_2(g) + 18 H_2O(l)
\][/tex]

What mass of octane must be burned in air in order to release 24 moles of [tex]\(CO_2\)[/tex] into the atmosphere?



Answer :

To determine the mass of octane (C_8H_{18}) needed to produce 24 moles of CO_2 through combustion, let's break down the problem into clear, systematic steps:

1. Write down the balanced chemical equation:
[tex]\[ 2 \, \text{C}_8\text{H}_{18} (l) + 25 \, \text{O}_2 (g) \rightarrow 16 \, \text{CO}_2 (g) + 18 \, \text{H}_2\text{O} (l) \][/tex]

2. Identify the molar ratio between C_8H_{18} and CO_2:
From the equation, we see that 2 moles of C_8H_{18} produce 16 moles of CO_2. Simplifying this ratio:
[tex]\[ \frac{2 \, \text{moles} \, \text{C}_8\text{H}_{18}}{16 \, \text{moles} \, \text{CO}_2} = \frac{1 \, \text{mole} \, \text{C}_8\text{H}_{18}}{8 \, \text{moles} \, \text{CO}_2} \][/tex]
Therefore, 1 mole of C_8H_{18} produces 8 moles of CO_2.

3. Determine the moles of C_8H_{18} required to produce 24 moles of CO_2:
[tex]\[ \frac{24 \, \text{moles} \, \text{CO}_2}{8 \, \text{moles} \, \text{CO}_2 / \text{mole} \, \text{C}_8\text{H}_{18}} = 3 \, \text{moles} \, \text{C}_8\text{H}_{18} \][/tex]

4. Calculate the molar mass of C_8H_{18}:
To find the molar mass, sum the masses of all atoms in the molecule:
- Carbon (C): [tex]\(8 \, \text{atoms} \times 12.01 \, \text{g/mol} = 96.08 \, \text{g/mol}\)[/tex]
- Hydrogen (H): [tex]\(18 \, \text{atoms} \times 1.008 \, \text{g/mol} = 18.144 \, \text{g/mol}\)[/tex]
- Therefore, molar mass of C_8H_{18} = [tex]\(96.08 \, \text{g/mol} + 18.144 \, \text{g/mol} = 114.224 \, \text{g/mol}\)[/tex]

5. Calculate the mass of C_8H_{18} required:
[tex]\[ 3 \, \text{moles} \, \text{C}_8\text{H}_{18} \times 114.224 \, \text{g/mol} = 342.672 \, \text{g} \][/tex]

So, the mass of octane that must be burned to produce 24 moles of CO_2 is approximately 342.672 grams.