Question

Limiting reactants Liquid octane (CH,(CH) CH,) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO,) and
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Balanced equation:

2 C8H18 + 25 O2 \rightarrow 16 CO2 + 18 H2O

Moles of Ocatane = mass / molar mass = 18.3 / 114 = 0.161 moles

Moles of O2 mass / molar mass = 84 / 32 = 2.625 moles

From the equation, mole sof O2 needed = moles of octane * 25 / 2 = 0.161 * 25 / 2 = 2.012 moles

Since there is more oxygen than needed, octane will completely react meaning miniumum leftover mass = 0 g

Add a comment
Know the answer?
Add Answer to:
Limiting reactants Liquid octane (CH,(CH) CH,) will react with gaseous oxygen (0) to produce gaseous carbon...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT