Question

explanation needed...... plz don't write something useless, i will report and that will be waste of time for both of us.....​

Answer 1

Nitrogen and Hydrogen combine to form Ammonia according to the following reaction:

$N_{2} + 3H_{2}→ 2NH_{3}$

From the above reaction, we can see that,

• 2 moles of N combines with (3×2), 6 moles of Hydrogen.
• The product formed is 2 moles NH3.

1 mole of N contains 14g of nitrogen.

1 mole of H contains 1g of Hydrogen.

Hence, from unitary method we can find that,

28g of Nitrogen combines with 6 grams

of Hydrogen.

Therefore, 56 gram of Nitrogen will combine with

$→\frac{ 56×6}{28} =12$

12g of Hydrogen, and 68g of Ammonia gets formed.

[ N (56g) + H (12g) = NH3 (68g) ]

(Rest Hydrogen remains as it is in the container, because the nitrogen will only react with 12g of Hydrogen.)

Here, Nitrogen acts as Limiting Reagent because Hydrogen is present in sufficient amount but nitrogen is limited.