Answer to Question #287564 in General Chemistry for Zaza

Question #287564

How many liters of O2 at STP is needed to completely burn 20.0 L of butane(C4H10), with this reaction?


2C4H10+13O2 ---> 8CO2+10H2O


1
Expert's answer
2022-01-17T15:24:02-0500

Solution:


given chemical reaction:


2C4H10+13O28CO2+10H2O2C_4H_{10}+13O_2 \to 8CO_2 +10 H_2O


I) We need to divide given volume of butane in order to find moles of it:


n=2022.4=0.8928moles  of  butanen=\frac{20}{22.4}=0.8928moles\;of\;butane


nO2=0.8928×132=5.799  moles  of  oxygenn_{O_2}=\frac{0.8928 \times 13}{2}=5.799\;moles \;of\;oxygen


II) We need to multiply moles of oxygen by volume in standard conditions


VO2=n×Vo=5.799×22.4LV_{O_2}=n \times V _o=5.799 \times 22.4 L


VO2=129.91LV_{O_2}=129.91L



Answer: We need 129.91 L of oxygen to completely burn 20.0 L of butane.





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Zaza
19.01.22, 09:08

Thank you so much for the answer.❤️ I really need this right now. Thank you ❤️❤️

Zaza
18.01.22, 01:35

Omg! Thank you so much for answering this.❤️ I really need it right now because I can't understand our lesson. Thank you.❤️

Leave a comment