Compare the stability of primary, secondary and tertiary carbocation
1 answer:
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Answer:
0.267 mol
Explanation:
The <em>unbalanced equation</em> for the reaction that takes place is:
- C₈H₁₈ + O₂ → H₂O + CO₂
Once we balance it we're left with:
- 2C₈H₁₈ + 25O₂ → 18H₂O + 16CO₂
Using the <em>stoichiometric ratio </em>of water and octane from the balanced reaction, we can <u>convert mol of water into mol of octane</u>:
- 2.40 mol H₂O *
= 0.267 mol C₈H₁₈
Thus 0.267 moles of octane are needed to produce 2.40 mol of water.