Chapter 11, Problem 43A

Interpretation Introduction

Interpretation:

The mole ratio for the equation needs to be determined.

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactants, which is present in limited amount and determines the amount of product is known as limiting reactant. Whereas another reactant is known as excess reactant.

Answer to Problem 43A

$\begin{array}{l}{\text{4 M}}_{\text{2}}\text{N }\stackrel{}{\to }{\text{ 2 M}}_{\text{4}}{\text{+ 2 N}}_{\text{2}}\\ \frac{{\text{4 moles M}}_{\text{2}}\text{N}}{{\text{2moles M}}_{\text{4}}}\text{ = }\frac{{\text{2 moles M}}_{\text{2}}\text{N}}{{\text{1 mole M}}_{\text{4}}}\\ \frac{{\text{4 moles M}}_{\text{2}}\text{N}}{{\text{2moles N}}_{\text{2}}}\text{ = }\frac{{\text{2 moles M}}_{\text{2}}\text{N}}{{\text{1 mole N}}_{\text{2}}}\end{array}$

Explanation of Solution

As per given figure in the square in the image, the balance chemical equation can write as:

${\text{4 M}}_{\text{2}}\text{N }\stackrel{}{\to }{\text{ 2 M}}_{\text{4}}{\text{+ 2 N}}_{\text{2}}$

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation. Thus, one mole ratio relates with other mole ratio. The mole ratio in the given balance reaction must be:

$\begin{array}{l}{\text{4 M}}_{\text{2}}\text{N }\stackrel{}{\to }{\text{ 2 M}}_{\text{4}}{\text{+ 2 N}}_{\text{2}}\\ \frac{{\text{4 moles M}}_{\text{2}}\text{N}}{{\text{2moles M}}_{\text{4}}}\text{ = }\frac{{\text{2 moles M}}_{\text{2}}\text{N}}{{\text{1 mole M}}_{\text{4}}}\\ \frac{{\text{4 moles M}}_{\text{2}}\text{N}}{{\text{2moles N}}_{\text{2}}}\text{ = }\frac{{\text{2 moles M}}_{\text{2}}\text{N}}{{\text{1 mole N}}_{\text{2}}}\end{array}$

Conclusion

Mole ratio can calculate with the help of balance chemical reaction equation.