Chapter 12.3, Problem 26SP

Interpretation Introduction

Interpretation: The limiting reagent is to be determined.

Concept Introduction: According to the law of conservation of mass, the mass can neither be created nor be destroyed. Every reaction must be balanced. With the help of the balanced equation, the limiting reagent can be found and in accordance with it, the amount of product can be calculated.

Answer to Problem 26SP

The limiting reagent is ${\text{O}}_{\text{2}}$ .

Explanation of Solution

Given:

• 2.70 mol ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$
• 6.30 mol ${\text{O}}_{\text{2}}$

The reaction is given as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

According to the balanced equation,

  $\begin{array}{c}1\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 3\text{mol}\text{of}{\text{O}}_{\text{2}}\\ 2.70\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 3\times 2.70\text{mol}\text{of}{\text{O}}_{\text{2}}\\ \to 8.1\text{mol}\text{of}{\text{O}}_{\text{2}}\end{array}$

The mol of ${\text{O}}_{\text{2}}$ is given as 6.30 mol means that it is present in lesser amounts. So, the limiting reagent is ${\text{O}}_{\text{2}}$ .