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FISH0818 FISH0818
wrote...
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12 years ago
Ammonia,   NH3,   is produced from the reaction of nitrogen and hydrogen gases.
In one experiment   2.38 g   of   H2   is mixed with   13.25 g   of   N2.

What is the theoretical yield of   NH3 ?   I.E., how many grams of   NH3   should be formed?
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wrote...
12 years ago
N2 + 3H2 -> 2NH3

N2 has a gfm of 28g/mol and H2 has a gfm of 2g/mol.  (you do the math to find the moles I will make a rough guess) you've got about .5 mols of N2 and about 1 mole of H2, so H2 is your limiting reagent.  your ratio of H2 to NH3 is 2molNH3/3moleH2 and you multiply that by the one mole of H2.

NH3's gfm is 17g/mol so .667molNH3 x 17g/molNH3 ~ 12g

*edit* make sure you do the math over because I was guessing and estimating where you should not *edit*
wrote...
12 years ago
the reaction would read:
3 H2 + N2 = 2 NH3
First you need to determine the limiting factor (do you have more N2 or more H2, but remember that for every mole of N2 your need 3 moles of H2).

H2 has a molecular weight of 2.002 and N2 has a molecular weight of 14.02.  

Dividing the grams of each substance by it's molecular weight yields the # moles of that substance.

There are 1.19 moles of H2 and .945 moles of N2.  To use all of the N2 you would need 3 times that amount in H2 (2.84 moles) and you only have 1.19 moles.  Therefore H2 is your limiting agent.

For every 3 moles of H2 you get 2 moles of NH3.  So you take the number of moles of H2 and divide by 3 and multiply by 2.  You get .79 moles of NH3.  Multiplying this by its molecular weight (17.02) gives you a yield of 13.47 grams.

Hope this helps!
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