Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: magicalatom on January 24, 2016, 09:04:05 PM
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Hi I need help with this problem below:
The question asks in acidic conditions, the sum of the coefficients in the products of the balanced reaction is:
MnO4-+C3H7OH---->Mn+2+C2H5COOH
I think this is a redox reaction balancing problem.
So far
I got 5e-+8H++MnO4- ------> Mn+2 + 4H2O
I can't seem to figure out how to do this with C3H7OH and then multiply each equation to help cancel out terms.
The answer is D. 20 but I don't know how to get this answer.
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C3H8O + H2O = C3H6O2 +.
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What is this and how do you arrive at 20 as the answer? Did you even read the problem?
C3H8O + H2O = C3H6O2 +.
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Just finish this reaction and add properly two half reactions!
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That is where I am having trouble. I got 6H+ + H2O + 2C3H7OH ---> 3C2H5COOH + 6e-
then I tried multiplying first equation by 6 and this second one by 5 so I could sum and cancel 30e-, but still can't get 20 on products side as coefficients?
Just finish this reaction and add proprly two half reactions!
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H+ should be on the right side of equation!
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Right side has 16 H's so the left side needed 6 more. Why do you say 6H's should be on the right side?
I looked up the rules on this. First you deal with metals then with H and O at the end.
I just seem to not see if propanol is able to split up ionically. I know it has a OH and H can come off.
H+ should be on the right side of equation!
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Check charge and material balance of your equation.
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I am still lost. I have been on this problem for 3 hours. I have to move on sorry.
Check charge and material balance of your equation.
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My hint reaction have correct coefficients on the left side. Nothing more is needed on this side.
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In your hint what happen to C3H7OH?
My hint reaction have correct coefficients on the left side. Nothing more is needed on this side.
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C3H7OH = C3H8O just for easier addition and subtraction
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Just learned another trick. Ok so I have
H2O + C3H8O ----> C3H6O2 + 2H+ + 1e- is this correct?
C3H7OH = C3H8O just for easier addition and subtraction
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Still wrong - both charges and hydrogens
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Oops I see
H2O + C3H8O ----> C3H6O2 + 4H+ + 4e- how about now?
Still wrong - both charges and hydrogens
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At last. Now use correct coefficient for both reactions to cancel electrons, add both reactions and move H+ to the left side
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Multiply top by 4 and second one by 5 to cancel 20e-? Then do I multiply again by something to cancel out more?
At last. Now use correct coefficient for both reactions to cancel electrons, add both reactions and move H+ to the left side
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After multiplying add both reactions by sides and calculate H+ on the left side and water on the right one
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I got 18. I first multiplied the second by 2 to get 8H+ and cancel those out. Then I did the new second equation by 2 to get 4H2O and cancel those out.
Then I had to divide by 2 to get 8e- and multiply the first one by 8 and the second one by 5 to cancel out 40e-
I ended up with products side coeffiecients as 18. Do you think the book made an error? No text is perfect. I have caught some errors in it.
After multiplying add both reactions by sides and calculate H+ on the left side and water on the right one
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I got 4, 5 and 11. Check your calculations
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Are you canceling the water too or just e- and H+?
I got 4, 5 and 11. Check your calculations
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Electrons should be canceled. All H+ on the left side, all H2O on the right side
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we have 5e- and 4e- so you use common factor 20 correct then cancel?
Electrons should be canceled. All H+ on the left side, all H2O on the right side
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Write down the final equation with electrons canceled and check charge balance and material balance
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I have
1. 5e- + 8H+ + MnO4- ---> Mn+2 + 4H2O
2. H2O + C3H8O ----> C3H6O2 + 4H+ + 4e- is this correct so far?
Write down the final equation with electrons canceled and check charge balance and material balance
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both equations are correct
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So I multiplied the top by 4 and the bottom one by 5 to cancel out 20e-
Is that correct first step?
both equations are correct
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Yes, and at this moment you can find 16-5 = 11H2O on the right side
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Nailed it BAAM :)
12H+ + 5C3H8O + 4MnO4- ----> 11 H2O + 4Mn+2 + 5 C3H6O2
So I multiplied the top by 4 and the bottom one by 5 to cancel out 20e-
Is that correct first step?
both equations are correct
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Well worth the 4 hours spent. Every moment taught me something. This is what I try to tell other students what studying is. Some will just never get it.
And thank you for not just showing me, but guiding me. Brain grows new pathways. It is like weightlifting. One must do problems mentally to grow brain power :)
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All my students should solve almost all redox reaction within 5 minutes time limit.
and some of these have max 1/2 hour limit
http://www.kchn.pg.gda.pl/pliki_forum/67d934b0a70cc7beaa91058b2fac48ec.pdf
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Thank you for the additional problems. I will practice. :)