To determine the number of moles of [tex]\( O_2 \)[/tex] needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex], we need to use stoichiometry based on the given balanced chemical equation:
[tex]\[
2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O
\][/tex]
The stoichiometric relationship between [tex]\( C_2H_6 \)[/tex] and [tex]\( O_2 \)[/tex] from the equation tells us that 2 moles of [tex]\( C_2H_6 \)[/tex] react with 7 moles of [tex]\( O_2 \)[/tex]. This ratio can be used to set up a proportion to find out how many moles of [tex]\( O_2 \)[/tex] are needed for 24 moles of [tex]\( C_2H_6 \)[/tex].
So, the set up is as follows:
[tex]\[
\frac{24 \text{ mol } C_2H_6}{2 \text{ mol } C_2H_6} = \frac{x \text{ mol } O_2}{7 \text{ mol } O_2}
\][/tex]
We solve for [tex]\( x \)[/tex], which represents the moles of [tex]\( O_2 \)[/tex] needed:
[tex]\[
\frac{24}{2} = \frac{x}{7}
\][/tex]
Cross-multiplying to solve for [tex]\( x \)[/tex]:
[tex]\[
24 \times 7 = 2 \times x
\][/tex]
[tex]\[
168 = 2x
\][/tex]
[tex]\[
x = \frac{168}{2}
\][/tex]
[tex]\[
x = 84
\][/tex]
Thus, 84 moles of [tex]\( O_2 \)[/tex] are needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex].
The correct setup from the given choices is:
[tex]\[
\begin{array}{rl}
24 \text{ mol } C_2H_6 & 7 \text{ mol } O_2 \\
& 2 \text{ mol } C_2H_6
\end{array}
\][/tex]
This setup properly reflects the stoichiometric conversion needed for the calculation.
