Chapter 9, Problem 31A

Interpretation Introduction

Interpretation: The representation of product mixture for the given chemical reaction of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ to form ${\text{NH}}_3$ molecules needs to be explained.

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant. It is used to determine the amount of product formed during the chemical reaction.

Answer to Problem 31A

  

Explanation of Solution

In the given chemical reaction, there are 6 ${\text{N}}_{\text{2}}$ molecules and 6 ${\text{H}}_{\text{2}}$ molecules as reactant molecules. The balance chemical reaction of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ to form ${\text{NH}}_3$ molecules can be written as:

  ${\text{N}}_{\text{2}}{\text{ + 3H}}_{\text{2}}\stackrel{}{\to }{\text{2NH}}_3$

Thus 1 mole of ${\text{N}}_{\text{2}}$ reacts = 3 moles of ${\text{H}}_{\text{2}}$

• Determine the limiting reactant:

  ${\text{6 mole H}}_{\text{2}}\times \frac{{\text{1 mole N}}_{\text{2}}}{{\text{3 moles H}}_{\text{2}}}{\text{= 2 molesN}}_{\text{2}}$

Since 6 moles of ${\text{N}}_{\text{2}}$ are given thus ${\text{N}}_{\text{2}}$ must be excess reactant and ${\text{H}}_{\text{2}}$ must be limiting reactant which will consume 2 moles of ${\text{N}}_{\text{2}}$ and 4 moles of ${\text{N}}_{\text{2}}$ will remain in mixture.

3 moles of ${\text{H}}_{\text{2}}$ = 2 moles ${\text{NH}}_3$

  ${\text{6 moles H}}_{\text{2}}\text{ × }\frac{{\text{2 moles NH}}_{\text{3}}}{{\text{3 moles H}}_{\text{2}}}{\text{=4 moles of NH}}_{\text{3}}$

Thus the representation must be:

  

Conclusion