To determine how many moles of [tex]\( O_2 \)[/tex] are needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex], we will use stoichiometry based on the provided balanced chemical equation:
[tex]\[ 2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O \][/tex]
1. Identify the stoichiometric ratio:
From the balanced equation, we see that 2 moles of [tex]\( C_2H_6 \)[/tex] react with 7 moles of [tex]\( O_2 \)[/tex].
2. Set up the proportion:
If 2 moles of [tex]\( C_2H_6 \)[/tex] require 7 moles of [tex]\( O_2 \)[/tex], then 24 moles of [tex]\( C_2H_6 \)[/tex] would need:
[tex]\[ \frac{7 \text{ moles of } O_2}{2 \text{ moles of } C_2H_6} \][/tex]
This represents the stoichiometric ratio.
3. Calculate the moles of [tex]\( O_2 \)[/tex] required for 24 moles of [tex]\( C_2H_6 \)[/tex]:
Using the proportion, we get:
[tex]\[ \text{Moles of } O_2 = \left( \frac{7 \text{ moles of } O_2}{2 \text{ moles of } C_2H_6} \right) \times 24 \text{ moles of } C_2H_6 \][/tex]
Solving this:
[tex]\[ \text{Moles of } O_2 = \left( \frac{7}{2} \right) \times 24 = 3.5 \times 24 = 84 \][/tex]
Therefore, 84.0 moles of [tex]\( O_2 \)[/tex] are needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex].
