[rookie]

I've been trying to solve this stoichiometric problem, but no matter what i do, it just won't come out.

N₂  +  4H₂O  ->  2H₂O₂ +  N₂H₄

- the question is how many grams of H₂O₂  are there in 24.32 g of N₂H₄

I first divided the 24.32g of N2H4 by its molar mass (32.052g) and then i multiplied it by the equivalence of  of H2O2 to N2H4 which came out to 2/1 then i multiplied it by the molar mass of H2O2 ... but my answers are coming out all wrong.. am i doing the problem correctly ?  

Modification: [sub ] and [/sub ] tags corrected.

[Chrataxe]

You have your <sub>[/ sub] backwards.  The one with the slash comes second.  

There are ZERO grams of H2O2 in N2H4...think about it.  I don't think that is what the question is asking.

Also, I'm curious about the reaction.  I don't know under what conditions nitrogen gas will react with water...BUT, since N2 is 78% of the gas in the atmosphere and Earth is about 75% water, we would have very little water on Earth, oceans would be made of hydrogen peroxide.  I know of a couple of ways to make hydrazine and this isn't one.  Not saying its not possible....just curious under what conditions this would happen.

I'm not really understanding what you are saying.  If I am tracking what you are saying, you divided the g of N2H4.  So, that number tells you the number of moles of N2H4.  Since it is a 2:1 ratio of H2O2:N2H4, the there would be twice as many moles of H2O2.  Multiply the number of moles of H2O2 by the atomic weight to give the number of grams.  From what I can tell, that is what you have done.  Double check your math....if you math is right, why do you think you have the wrong answer?</sub>

[Albert]

I totally agree with Chrataxe: I was taken aback when I read the question!

But, I'll try to read between the lines and find what Rookie might have wanted to say.

We have 24.32 g of N2H4. So: 2*14.007 = 28.014% of N in this molecule. This means we have 6.813g of N among the products and N2 was, in the beginning,  6.813g

1 mole of N2 reacts with 4 moles of H2O: we had 17.52g of water.

To sum up: (17.52 + 6.813)g = (24.32 + x)
X = H2O2 mass = 18.95 mg

Could it be?

[Chrataxe]

Where did you get the 6.813 from?  

(24.32)/32.06= moles of N2H4

H2O2:N2H4 = 2:1

2 x moles of N2H4 = # moles of H2O2

moles of H2O2 x GAW = g H2O2

My calculations give me 51.61g.  Since both have about the same GAW, and the ratio is 2:1, there should be about 2x as much H2O2.

[rookie]

ya i got 51.61 g too... i think maybe the answer in the book was given incorrectly

[Albert]

Where did you get the 6.813 from?  

28.014% : x g of N = 100% : 24.32 g

6.813g is the weight of nitrogen in 24.32g of N2H4. So it must also be the mass of nitrogen among the reagents.

[Borek]

See the picture

[Chrataxe]

So I see.  But, that still poses the question of what that had to do with the question....which is what confused me in the first place.

[Borek]

On the serious note - you were right from the beginning. Reaction seems impossible and the question in original wording doesn't make sense (or rather the only answer is 0.0).

[mike]

Looks to me like the reaction is back to front, should be hydrazine + peroxide maybe?

Isn't this some kind of rocket fuel?