Chapter 13, Problem 100A

Interpretation Introduction

Interpretation: Limiting reagent in the given reaction is to be identified.

Concept introduction:

A reactant that is completely consumed at the end of a chemical reaction is known as the limiting reagent. Since the reaction cannot proceed without this reagent, the amount of product that can be produced is constrained.

Answer to Problem 100A

  ${\text{C}}_{\text{2}}{\text{H}}_4$

Explanation of Solution

The given reaction is as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}+{\text{H}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_6$

Amount of ${\text{C}}_{\text{2}}{\text{H}}_4$ is 40.0 g.

Amount of ${\text{H}}_{\text{2}}$ is 3.0 g.

Molar mass of ${\text{H}}_{\text{2}}$ is $\text{2 }\text{g}/\text{mol}$ .

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_4$ is $28\text{g}/\text{mol}$ .

Now, calculate the molar masses of ${\text{C}}_{\text{2}}{\text{H}}_4$ and ${\text{H}}_{\text{2}}$ .

  $\begin{array}{c}{\text{molar mass of H}}_{\text{2}}=\frac{\text{mass in g}}{\text{molar mass}}\\ =\frac{3.0}{2.0\text{g}/\text{mol}}\\ =\text{1}\text{.5 mol}\end{array}$

  $\begin{array}{c}{\text{molar mass of C}}_{\text{2}}{\text{H}}_4=\frac{\text{mass in g}}{\text{molar mass}}\\ =\frac{40.0}{\text{ 28}\text{g}/\text{mol}}\\ =\text{1}\text{.42 mol}\end{array}$

From the above reaction, 1 mol of ${\text{C}}_{\text{2}}{\text{H}}_4$ reacts with 1 mol of ${\text{H}}_{\text{2}}$ . Here, no. of moles of ${\text{C}}_{\text{2}}{\text{H}}_4$ is less. Thus it is a limiting reagent.