Question
If you have 14.2 mol of C4H10 and 85.6 mol of O2, which is the limiting reagent?
I calculated mass using m=N*M, giving me the answer that C4H10 is the limiting reagent. That being said, the answer on the back says O2.
I calculated mass using m=N*M, giving me the answer that C4H10 is the limiting reagent. That being said, the answer on the back says O2.
Answers
Anonymous
Write a balance equation first.
C4H10 + O2 ---> CO2 + H2O
Balance the reaction:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O
So, you need 13 moles of O2 for 2 moles of C4H10.
moles of O2=85.6/13=6.58
moles of C4H10=14.2/2=7
Moles of O2 is the limiting reagent.
C4H10 + O2 ---> CO2 + H2O
Balance the reaction:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O
So, you need 13 moles of O2 for 2 moles of C4H10.
moles of O2=85.6/13=6.58
moles of C4H10=14.2/2=7
Moles of O2 is the limiting reagent.