To determine how many moles of ethane (C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]) would react with 35 moles of oxygen (O[tex]\(_2\)[/tex]), let's analyze the balanced chemical equation:
[tex]\[
2 \, \text{C}_2\text{H}_6 + 7 \, \text{O}_2 \rightarrow 6 \, \text{H}_2\text{O} + 4 \, \text{CO}_2
\][/tex]
From this equation, we see the stoichiometric coefficients that show how many moles of each reactant and product are involved in the reaction:
- 2 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] react with 7 moles of O[tex]\(_2\)[/tex].
Given that we have 35 moles of O[tex]\(_2\)[/tex], we need to find the corresponding moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]. The stoichiometric ratio between C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] and O[tex]\(_2\)[/tex] is:
[tex]\[
\frac{2 \, \text{moles of C}_2\text{H}_6}{7 \, \text{moles of O}_2}
\][/tex]
Using this ratio, we can calculate the moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] needed for 35 moles of O[tex]\(_2\)[/tex]:
[tex]\[
\text{moles of C}_2\text{H}_6 = \left( \frac{2}{7} \right) \times 35
\][/tex]
Simplifying the calculation:
[tex]\[
\text{moles of C}_2\text{H}_6 = \left( \frac{2 \times 35}{7} \right) = \left( \frac{70}{7} \right) = 10
\][/tex]
Therefore, the number of moles of ethane (C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]) that would react with 35 moles of O[tex]\(_2\)[/tex] is 10.
So, the correct answer is:
B. 10
