Question

Bonjour!



Q. 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3(g). Calculate the moles of NH3(g) formed. Identify the limiting reagent in the production of NH3 in this situation.

#Answer the question with steps#

Answer 1

Hope you get the answer and adjust with handwriting

Answer 2

=====================
HERE IS YOUR ANSWER ☞
=====================

The balanced chemical equation for formation of ammonia is

[ N2 + 3H2 ===> 2NH3. ]

● molecular mass of N2 = 28 gm/mol
= 0.028 kg/mol

● molecular mass of H2 , = 6 gm/mol
= 0.006 kg/mol

● molecular mass of NH2 = 17 gm/mol
= 0.017 kg/mol

Now,
according to the balanced chemical equation

.
,
0.028 kg of N2 reacts with 0.006 kg of H2

so,

● 50 kg of N2 react with =>

=(0.006×50)/0.028

= 10.71 kg of H2

the amount of H2 given (10. kg) is less then the amount required (10.71 kg) for 50kg of N2.

therefore,

=======
H2 is the limiting reagent
=======

● the formation of NH3 will depend on the amount of H2 available for reaction 0.006kg of N2 produce =>

=(2×0.017)

= 0.034 kg of NH3

so,

10 kg of H2 will produce =>

= (0.034×10)/0.006

= 56.67 kg of NH3 ______ANSWER

=====================
THANK YOU ☺☺☺☺
=====================

DEVIL_KING ▄︻̷̿┻̿═━一