Question

50kg N₂(g) and lokg of H₂ (g) are mixed to
produce NH3(g...calculate the amount of NH3-
(g) formed Identify the limiting reagent
the production of NH₃ in this solution
N₂(g) + 3H₂ (g)=2NH₃(g).​

Answer 1

Answer:

H2 is the limiting reagent  and the amount of NH3 formed is 56.6 Kg

Explanation:N2 + 3H2 = 2NH3

According to this formula,

28 Kg N2 + 6 Kg H2 combines to form 34 Kg

That is ,

1 kg N2 reacts with 6/28 Kg H2 to form 34/28 Kg NH3

Therefore.

50 Kg N2 reacts with (6*50)/28 =10.7 Kg NH3

But we have only 10 kg H2. There for H2 is the liming reagent.

So.

28 Kg N2 + 6 Kg H2 combines to form 34 Kg

 

28/6 Kg N2 + 1 Kg H2 to form 34/6 Kg NH3

Therefore

(28*10)/6 =46.6 Kg N2 + 10 Kg H2 reacts to form

(34*10)/6=56.6  Kg NH3