stoichiometry problem?

[PdiM]

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Hey,

I had a problem with the following question from a Kaplan simulated exam. Maybe somebody would be able to help me out:

given
HNO3 --> HNO2 --> H2N2O2 --> N2O(g) --> N2 (g)

If all the oxygen in the nitric acid is converted to water, how many additional equivalents of acid will be consumed during the production of 5M of nitrogen?

a) 20 b) 30 c) 40 d) 50

i've been stratching my head with this one, but can't seem to figure it out. I havent taken a look at kaplan's explanation yet. any takers??

 

[Mudd]

My assumption is that the arrows represent intermediates, which don't matter much in a stoichometry question. Assuming the overall reaction is:

HNO3 ---> H2O + N2

Then the balanced version, excluding hydrogens, is:

2 HNO3 ---> 6 H2O + 1 N2

There is a shortage of 10 H atoms on the reactant side of the equation. To generate 5 moles of N2 would require an additional 50 moles of protons from some source.

Something is wrong with this problem however, because from HNO3 to N2 is reduction, and there is no oxidation half-reaction shown. Is there more to the question?

 

[PdiM]

the answer IS 50 🙂

wow i was approaching that question from a totally wrong perspective - i thought when they were asking for "acid equivalents" they meant additional HNO3.

This was all that was stated in the question. Does it HAVE to be accompanied with the oxidation half-reaction? they couldnt just show the reduction half-reaction?

 

[Tweetie_bird]

aaww you little teaser
Didn't even give the answer!

Anyway, is it "B"....30 extra acid equivalents? Lemme know before the curiosity kills me. 😀

 

[PdiM]

hehe i thought it would drive people to actually work out the problem if i didnt give the answer 😉

its d) 50

 

[Mudd]

You are correct, there is no need to specify an oxidation half-reaction, but to balance the reaction, they need 10 e- on the reactant side for every N2 that is formed. This would have allowed you to balance the reaction by charge, rather than stoichiomtery. Seeing 10 e- would have made it more clear that 10 H+ were required, or that 10 HNO3 would provide the 10 H and yield 10 nitrate anions in the product of the reduction half-reaction.