Problem: Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia.N2(g)+3 H2(g) → 2NH3(g)Assume 4 molecules of nitrogen and 9 molecules of hydrogen are present.1.) After complete reaction, how many molecules of ammonia are produced?2.How many molecules of H2 remain?3.How many molecules of N2 remain?4. What is the limiting reactant, Hydrogen or Nitrogen?

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N2(g) + 3 H2(g) → 2 NH3(g)

Notice that we are given the mass of both reactants: this means we need to determine the limiting reactant, which is the reactant that forms the less amount of product


This is because once the limiting reactant is all used up, the reaction can no longer proceed and make more products.

This means the limiting reactant determines the maximum mass of the product (H2O) formed.


We need to perform a mole-to-mole comparison between each reactant and  NH3


4 molecules of nitrogen and 9 molecules of hydrogen are present

4 molecules N2 ×2 molecules NH31 molecules N2= 8 molecules NH39 molecules H2 ×2 molecules NH33 molecules H2= 6 molecules NH3

Since H2 produced the least reactant, it is the limiting reagent. 

1. After complete reaction, there are 6 molecules of NH3.

Since H2 is the limiting reactant, there will be no H2 left. 


2. 0 molecules of H2 remain


Calculate moles N2 used:

9 molecules H2 ×1 molecules N23 molecules H2= 3 molecules N2


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Problem Details

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

N2(g)+3 H2(g) → 2NH3(g)

Assume 4 molecules of nitrogen and 9 molecules of hydrogen are present.

1.) After complete reaction, how many molecules of ammonia are produced?

2.How many molecules of H2 remain?

3.How many molecules of Nremain?

4. What is the limiting reactant, Hydrogen or Nitrogen?

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