Chapter 9, Problem 66A

Interpretation Introduction

Interpretation: The correct option which gives the least amount of product for the given reaction needs to be determined.

  ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

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.

Answer to Problem 66A

Correct answer: Option (b) is the correct answer.

Explanation of Solution

Reason for correct option: In the given reaction; the ratio of ${\text{H}}_{\text{2}}$ and ${\text{O}}_{\text{2(g)}}$ is 2 : 1 thus the reactant which is present in lesser amount will be limiting reactant.

  ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

In option (b) there are 8 moles of ${\text{H}}_{\text{2}}$ with 2 moles of ${\text{O}}_{\text{2(g)}}$ . So according to 2 : 1 mole ration ${\text{O}}_{\text{2(g)}}$ must be limiting reactant and the amount of product ${\text{H}}_{\text{2}}\text{O}$ formed must be:

  ${\text{2 moles O}}_{\text{2}}\text{ ×}\frac{{\text{2 moles H}}_{\text{2}}\text{O}}{{\text{1mole O}}_{\text{2}}}{\text{= 4 moles H}}_{\text{2}}\text{O}$

Conclusion

Reasons for incorrect options: In option a and d; the ratio of ${\text{H}}_{\text{2}}$ and ${\text{O}}_{\text{2(g)}}$ is 2 : 1 thus there is no limiting reactant.

  $\begin{array}{l}{\text{a) 3 moles O}}_{\text{2}}\text{ ×}\frac{{\text{2 moles H}}_{\text{2}}\text{O}}{{\text{1mole O}}_{\text{2}}}{\text{= 6 moles H}}_{\text{2}}\text{O}\\ {\text{d) 4 moles O}}_{\text{2}}\text{ ×}\frac{{\text{2 moles H}}_{\text{2}}\text{O}}{{\text{1mole O}}_{\text{2}}}{\text{= 8 moles H}}_{\text{2}}\text{O}\end{array}$

In option (c) ${\text{H}}_{\text{2}}$ must be limiting reactant as it is present in lesser amount.

  ${\text{ 5 moles H}}_{\text{2}}\text{ ×}\frac{{\text{2 moles H}}_{\text{2}}\text{O}}{{\text{2 mole H}}_{\text{2}}}{\text{= 5 moles H}}_{\text{2}}\text{O}$

All options are making different amount of products thus option (e) is also incorrect.