Question

14g of N2 reacts with 6g of H2 to form ammonia. 1. Identify the limiting reagent 2. calculate the amount of reagent which remains unreacted. 3. calculate the amount of ammonia produced?

Answer 1

N2+3H2=2NH3
28gm. 6 gm. 34 gm
28 gm N2 require 6g.m
14g.m will require 3g.m H2
3 gm of H2 will remain unreacted
so N2 is limiting reagent in rxn
N2 will. decide the formation of NH3
28gm of N2 produce = 34 gm
14 g.m. will produce=17 gm NH3
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limiting reagent is N2
Amount of NH3 produce =17gm

Answer 2

The balanced Chemical equation of the process is given below :
N2 + 3H2 → 2NH3
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We note from the above equation that 3mol (or 3×2 = 6g) of H2 is required with 1 mol (or 1×28=28g) of N2 to produce 2 moles (or 2×17=34g) of NH3.
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#1) IDENTIFYING LIMITING REAGENT:

Given :
N2 = 14 g .
H2 = 6 g .
we know that ;
28g N2 requires → 6g H2
=> 1g N2 requires → 6/28 g H2
=> 14 g N2 requires → (6/28)×14 g H2
=> 14g N2 requires → 3 g of H2 .
But we are given "6g " of H2 so H2 is in excess ( not a limiting reagent).
But we have only "14g" of N2 that is insufficient to react completely with "6g" of H2 . Hence N2 is LIMITING REAGENT .
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#2) AMOUNT OF EXCESS REAGENT LEFT AFTER REACTION :

We know that N2 (14g,given) will be completely used up and a part of the given amount of H2 (=3g , calculated above in #1) required in reaction is also used up in the reaction .
so H2 left unreacted = 6-3 = 3 g .
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#3) AMOUNT OF AMMONIA PRODUCED:

We know N2 is LIMITING here . So Amount of NH3 produced will depend only on N2 amount .
From the Chemical equation we note that :
28g N2 produces → 34g of NH3
=> 1g of N2 produces → 34/28 g of NH3
=> 14 g of N2 produces → (34/28)×14g of NH3 .
=> 14 g of N2 produces → 17 g of NH3 .
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hope it helps !