Chapter 11, Problem 42A

Interpretation Introduction

Interpretation:

The number of mole ratio the given reaction with respect to ammonium dichromate to the product needs to be determined.

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\stackrel{}{\to }{\text{ N}}_{\text{2}}{\text{ + Cr}}_{\text{2}}{\text{O}}_{\text{3}}{\text{ + 4 H}}_{\text{2}}\text{O}$

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 42A

$\begin{array}{}$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{ N}}_{\text{2}}\text{ = 1 : 1 }$$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{ = 1 : 1}$$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{ 4 H}}_{\text{2}}\text{O = 1: 4 }$\end{array}$

Explanation of Solution

The given reaction is as follows:

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\stackrel{}{\to }{\text{ N}}_{\text{2}}{\text{ + Cr}}_{\text{2}}{\text{O}}_{\text{3}}{\text{ + 4 H}}_{\text{2}}\text{O}$

According to the balance chemical reaction, 1 mole of ammonium chromate forms 1 mole of nitrogen, chromium (III) oxide and water. Thus the mole ratio must be:

$\begin{array}{}$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{ N}}_{\text{2}}\text{ = 1 : 1 }$$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{ = 1 : 1}$$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{ 4 H}}_{\text{2}}\text{O = 1: 4 }$\end{array}$

Conclusion

Thus,

$\begin{array}{}$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{ N}}_{\text{2}}\text{ = 1 : 1 }$$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{ = 1 : 1}$$ {{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}:{\text{ 4 H}}_{\text{2}}\text{O = 1: 4 }$\end{array}$