Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g)⟶2NH3(g) Assume 0.140 mol N2 and 0.459 mol H2 are present initially. After complete reaction, how many moles of ammonia are produced?

Answer 1

Answer:

moles of ammonia produced = 0.28 moles

Explanation:

The reaction is

$N_{2}(g)+3H_{2}(g) --> 2NH_{3}(g)$

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia

So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen

             = 0.42 moles of hydrogen molecule.

this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia

the moles of ammonia produced = 0.28 moles

Here the nitrogen is limiting reagent.