50.0kg of N2[g] and 10.0 kg of H2 [g] are mixed to produce NH3[g] formed.Identify the limiting reagent in production of NH3 in this situaton

 

Asked by virubloda6 | 21st May, 2019, 08:39: AM

Expert Answer:

Given:
 
Mass of N2 = 50 kg
 
                 = 50 × 103 gm
 
No. of moles of  N2
equals space fraction numerator 50 cross times 10 cubed over denominator 28 end fraction

equals 1.78 space cross times 10 cubed space moles
 
 
Mass of H2 = 10 kg
                 = 10 × 103 gm
 
No. of moles of  H2
 
equals fraction numerator 10 cross times 10 cubed over denominator 2 end fraction

equals 5 cross times space 10 cubed space mole
 
Formation of ammonia;
 
space space space straight N subscript 2 space space space plus space space 3 straight H subscript 2 space space rightwards arrow with space space space on top space 2 NH subscript 3

1 space mole space space space 3 space mole space space space space space space space space 2 space mole
 
From reaction, 1 mole of Nrequires 3 moles H2.
 
So 1.78× 103 mole of Nwill require  (1.78×103 ×3)  = 5.34 ×103 moles of H2.
 
Here 5×103 moles of H2 are given, therefore H2 is the limiting reagent.
 
 
Now  5×103 moles of H2 will give,
 
equals space fraction numerator 5 cross times 10 cubed cross times 2 over denominator 3 end fraction

equals space 3.33 space cross times 10 cubed space mole space of space NH subscript 3
 
Therfore the amount of NH3 produced is 3.33×103 mole.

Answered by Varsha | 21st May, 2019, 11:52: AM