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Need help with 2 multi chem problems: one is combustion, the other is redox equation

  • #1

Homework Statement


Really appriciate helps from anyone, these two probs hurt my head
1. Complete combustion of a 4.10g sample containing only propane (C3H5) and pentane (C5H12), yielded 12.42g CO2 and 6.35g H2O. What is the mass % propane in this sample?
a.4.50% b.38.0% c. 50.0% d.80.0% e.30.0%

2. Balance the following equation in basic solution:
Cr^(3+) + O2^(2-) = CrO4^(2-)
the sum of the coefficients is:
a.38 b.27 c.53 d.18 e.49

Homework Equations



C3H5 + (17/4)O2 = 3CO2 +(5/2)H2O
C5H12 + 8 O2 = 5CO2 + 6H2O

The Attempt at a Solution



For the first prob, i tried to put x is the mole of C3H5 and y is the mole of C5H12. Then i have sum of their mass equal 4.10g. On other part, i have the moles of CO2 from two equation as the factor of x and y, and then get the sum of them equal 12.42g. I solved the equation systems but get the x is negative? not sure what i did wrong? If anyone know plz tell me, otherwise i think the problem author gave the wrong information.
For the second prob, nothin i can do, i don't know if O2 is reduction or oxidation agent, don't even think O2^(2-) even exists, but my instructor seems pretty sure that equation can be balanced. Very thankful for the help from anyone.
 

Answers and Replies

  • #2
867
0


First you need one correct and balanced equation with integer coefficients. That propane needs to be C3H8
The left side will have propane and pentane with O2 and the equation will be much easier to work with.

Your first equation with x and y is good. For the other, convert that mass of CO2 to moles first so you're working with just moles. Then solve the system of equations and convert the moles x and y to masses to calculate the % by mass. Haven't done it myself but it looks like you're on the right track.

Can't help you with the other question, but O22- exists and it is the peroxide ion.
 
Last edited:
  • #3


propane is C3H8!!!! Damn! 2 hours hitting my head with this prob for nothin until now, i should have learned in heart those shameful, ridisculous organic names. thx bohrok.
i figured out balanced equation for #2 is 4(OH)^- + 2Cr^(3+) + 3O2^(2-) = 2CrO4^(2-) + 2H2O. but 4 + 2 + 3 + 2 + 2 = 13 for coefficient, not in multiple choices??? i guess the author this prob has some trouble with adding... or maybe just me -_-
If anyone points out the wrongness in my balance, very thankful!!!
 
  • #4
Borek
Mentor
28,327
2,714


What is O22-? Such ion is present in some ionic oxides, but not in solutions.

2Cr3+ + 3O22- + 4OH- -> 2CrO42- + 2H2O

is correct, and 13 it is.
 

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