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We want to find out how many moles of [tex]\(O_2\)[/tex] are needed to react with 24 moles of [tex]\(C_2H_6\)[/tex].

[tex]\[2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O\][/tex]

Which setup is correct?

[tex]\[
\begin{array}{l|l}
24 \, \text{mol} \, C_2H_6 & 2 \, \text{mol} \, C_2H_6 \\
& 7 \, \text{mol} \, O_2
\end{array}
\][/tex]

[tex]\[
\begin{array}{rl}
24 \, \text{mol} \, C_2H_6 & 7 \, \text{mol} \, O_2 \\
& 2 \, \text{mol} \, C_2H_6
\end{array}
\][/tex]



Answer :

GinnyAnswer
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.

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