As chem help

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Imsolost247
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Hi so i am currently studying yr12 chem and am revising for my mocks. Im going over reacting mass calculations and im so lost. Please can someone help. The question is:

In the manufacture of ammonia, what mass of ammonia can theoretically be formed when 1kg if nitrogen reacts with 0.5kg of hydrogen. N2 + 3H2 > 2NH3

I did 1/28 = N2 moles = 0.0357
Then 0.5/2 = 0.25 0.25/3=0.083 = H2 moles
So N2 is limiting so you do 0.0357x17= 0.607kg of 2NH3 ??
But i cant see where Im going wrong but my paper is saying the answer is 1210g ??

Please can someone explain, thanks
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username3718068
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(Original post by Imsolost247)
Hi so i am currently studying yr12 chem and am revising for my mocks. Im going over reacting mass calculations and im so lost. Please can someone help. The question is:

In the manufacture of ammonia, what mass of ammonia can theoretically be formed when 1kg if nitrogen reacts with 0.5kg of hydrogen. N2 + 3H2 > 2NH3

I did 1/28 = N2 moles = 0.0357
Then 0.5/2 = 0.25 0.25/3=0.083 = H2 moles
So N2 is limiting so you do 0.0357x17= 0.607kg of 2NH3 ??
But i cant see where Im going wrong but my paper is saying the answer is 1210g ??

Please can someone explain, thanks
Firstly, when dealing with calculating number of moles,
always use grams and not kg (leave the SI units for the physicists)

once you calculated # moles of each N2 and H2,
you concluded that N2 is limiting reagent, however don't multiply straight by 3. Remember that H2 is still reacting so you still have to take that into account before deciding what the limiting reagent is, if any exists.

Think, if all the N2 reacted, how much H2 is required? (hint: mol N2 x 3). Is this greater than the amount of H2 you already have? If so, then technically speaking the H2 is the limiting reagent, and therefore to calculate mol NH3 you simply look at the stochiometry coefficient of H2:NH3 then calculate the mass of NH3 from there.

If the amount of H2 is equal to amount of H2 you already have, then do there is no limiting reagent. Hence you can compare any stochiometry coeffiicient to then calculate mass of ammonia.

If however, the amount of H2 (to competely react with all N2 molecule) is less than what you already have, then yes you can conclude that N2 is the limiting reagent. Therefore you would continue to calculate mass of NH3 in the way would normally (in your method). If your final answer is wrong, then you will need to reconsider if there is a limiting reagent, and if so, what it is.
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Imsolost247
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#3
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Yes i tried it by changing it into g and follow what you said. Finally got the answer!!!! (even if it was to the wrong sig fig)
Thank youu😊
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