N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(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 (H _{2}O) formed**.

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

4 molecules of nitrogen and 9 molecules of hydrogen are present

$\mathbf{4}\mathbf{}\overline{)\mathbf{molecules}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}\mathbf{}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{molecules}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\phantom{\rule{0ex}{0ex}}\mathbf{9}\mathbf{}\overline{)\mathbf{molecules}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{3}\mathbf{}\overline{)\mathbf{molecules}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}$

Since H_{2} produced the least reactant, it is the limiting reagent.

**1. After complete reaction, there are 6 molecules of NH _{3}.**

Since H_{2} is the limiting reactant, there will be no H_{2} left.

**2. 0 molecules of H _{2} remain**

Calculate moles N_{2} used:

$\mathbf{9}\mathbf{}\overline{)\mathbf{molecules}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}{\mathbf{3}\mathbf{}\overline{)\mathbf{molecules}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{N}}_{\mathbf{2}}$

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

N_{2}(g)+3 H_{2}(g) → 2NH_{3}(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 H_{2} remain?

3.How many molecules of N_{2 }remain?

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

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