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Title: Ammonia, NH3, is produced from the reaction of nitrogen and hydrogen gases.? Post by: FISH0818 on Feb 9, 2012 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? Title: Ammonia, NH3, is produced from the reaction of nitrogen and hydrogen gases.? Post by: fire_crystal on Feb 9, 2012 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* Title: Ammonia, NH3, is produced from the reaction of nitrogen and hydrogen gases.? Post by: smhchre on Feb 9, 2012 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! |