# Homework Help: Chemistry: gases, exercise

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1. Mar 6, 2016

### ducmod

1. The problem statement, all variables and given/known data
Hello!
I would be grateful for your help on finding my mistake - somehow my answers are
wrong compared to the answer from the book.
Here is the exercise from a paragraph on gases (cumulative gas law, basics):

Consider the reaction for the synthesis of nitric acid:
3NO2 + H2O = 2HNO3 + NO
(a) If 12.8 L of NO2 measured at STP, is allowed
to react with 14.9 g of water, find the limiting
reagent and the theoretical yield of HNO3 in
grams.

Here is my solution:

1 mole of gas at STP = 22.4 L; 12.8 L gives 0.5714 moles of NO2
1 mole of H2O = 18.016 g; 14.9 g = 0.827 moles of H2O

The limiting reagent is NO2, because
3 moles NO2 + 2 moles of H2O
0.5714 moles NO2 requires only 0.381 moles of H2O, but we have 0.827 moles of H2O

1 mole of NO2 = 46.007 g
0.5714 moles of NO2 = 26.288 g
Theoretical yield of HNO3:
3 moles of NO2 = 2 moles of HNO3
0.5714 moles of NO2 = 0.3809 moles of HNO3

1 mole of HNO3 = 63.008
0.3809 moles of HNO3 = 24.001 g

Thank you!

2. Mar 6, 2016

### Staff: Mentor

Why 2 moles of H2O?
The conclusion is still right.

Looks right.
I don't think you have to work with decimal numbers of molar masses, they are very close to integers for the involved elements.

3. Mar 6, 2016

### Staff: Mentor

There is the mistake mfb already pointed out, but it doesn't influence the final answer.

What is the answer given in the book?

4. Mar 6, 2016

### ducmod

Thank you! Sorry, yes, only one mole of H2O.

5. Mar 6, 2016

### ducmod

I think the problem was in the question - there is some typo in the book, because exercise is asking about theoretical yield of HNO3,
while the book gives answer on grams of NO2