To solve this question, we need to understand the stoichiometric relationship between octane (C[tex]\(_8\)[/tex]H[tex]\(_{18}\)[/tex]) and carbon dioxide (CO[tex]\(_2\)[/tex]) in the combustion reaction. The balanced chemical equation for the combustion of octane is:
[tex]\[
2 \, \text{C}_8\text{H}_{18} + 25 \, \text{O}_2 \rightarrow 16 \, \text{CO}_2 + 18 \, \text{H}_2\text{O}
\][/tex]
From this balanced equation, we observe that 2 moles of C[tex]\(_8\)[/tex]H[tex]\(_{18}\)[/tex] produce 16 moles of CO[tex]\(_2\)[/tex]. To determine the moles of octane needed to produce 0.500 moles of CO[tex]\(_2\)[/tex], we use the stoichiometric relationship:
[tex]\[
\left( \frac{2 \, \text{moles C}_8\text{H}_{18}}{16 \, \text{moles CO}_2} \right)
\][/tex]
We set up the proportion to find the moles of octane required:
[tex]\[
\left( \frac{2 \, \text{moles C}_8\text{H}_{18}}{16 \, \text{moles CO}_2} \right) = \left( \frac{x \, \text{moles C}_8\text{H}_{18}}{0.500 \, \text{moles CO}_2} \right)
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = \left( \frac{2 \, \text{moles C}_8\text{H}_{18}}{16} \right) \times 0.500 \, \text{moles CO}_2 = 0.0625 \, \text{moles C}_8\text{H}_{18}
\][/tex]
Finally, we round this result to 3 significant digits:
[tex]\[
x = 0.062 \, \text{moles C}_8\text{H}_{18}
\][/tex]
Thus, the moles of octane needed to produce 0.500 mol of CO[tex]\(_2\)[/tex] is [tex]\( \boxed{0.062 \, \text{moles} \, \text{C}_8\text{H}_{18}} \)[/tex].
