Answer to Question #170198 in General Chemistry for josh

Question #170198

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.


N


2

(g)+3H


2

(g)⟶2NH


3

(g)

N2(g)+3H2(g)⟶2NH3(g)

Assume 0.280 mol N


2

0.280 mol N2 and 0.894 mol H


2

0.894 mol H2 are present initially.

After complete reaction, how many moles of ammonia are produced?


1
Expert's answer
2021-03-09T07:45:45-0500

Solution:


As is clear from the chemical equation, the molar ratio "\\frac{H_2} {N_2} =\\frac{3 }{1}".

"\\frac{n(H_2) } {n(N_2) } =\\frac{0.46 }{0.14}=3.286" i.e. H2 was taken in excess and some part of it remains unreacted. N2 s limiting reactant, it reacts completely and 0 moles of N2 remains unreacted.

Number of moles of NH2 produced may be calculated from the proportion:

 1 mol (N2) – 2 mol (NH3) (according to the chemical equation)

0.140 mol (N2) – X mol (NH3)

X = 0.140 · 2 / 1 = 0.280 mol of ammonia are produced.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS