Certainly! Let's solve this problem step-by-step using the given balanced chemical equation:
[tex]\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \][/tex]
1. Identify the molar ratios from the balanced equation:
- 1 mole of [tex]\( \text{CH}_4 \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex].
- 1 mole of [tex]\( \text{CH}_4 \)[/tex] also produces 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
2. Given that we have 6 moles of [tex]\( \text{H}_2\text{O} \)[/tex], we need to determine how many moles of [tex]\( \text{CO}_2 \)[/tex] this corresponds to.
3. Use the molar ratio between [tex]\( \text{H}_2\text{O} \)[/tex] and [tex]\( \text{CO}_2 \)[/tex]:
- According to the balanced equation, 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex] are produced for every 1 mole of [tex]\( \text{CO}_2 \)[/tex].
4. Set up the ratio to find the moles of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\frac{\text{moles of } \text{H}_2\text{O}}{\text{moles of } \text{CO}_2} = \frac{2}{1}
\][/tex]
5. Substitute the given moles of [tex]\( \text{H}_2\text{O} \)[/tex] into the ratio:
[tex]\[
\frac{6 \text{ moles of } \text{H}_2\text{O}}{\text{moles of } \text{CO}_2} = \frac{2}{1}
\][/tex]
6. Solve for the moles of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\frac{6 \text{ moles of } \text{H}_2\text{O}}{ \frac{2}{1}} = \text{moles of } \text{CO}_2
\][/tex]
Rearrange to solve for moles of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\text{moles of } \text{CO}_2 = \frac{6 \text{ moles of } \text{H}_2\text{O}}{2} = 3 \text{ moles of } \text{CO}_2
\][/tex]
7. Conclusion:
By following these steps, we determine that 3 moles of [tex]\( \text{CO}_2 \)[/tex] will be produced along with 6 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
Therefore, the correct answer is 3 moles of [tex]\( \text{CO}_2 \)[/tex].
