Chapter 12.3, Problem 28SP

(a)

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 28SP

The limiting reagent is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ .

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 as follows:

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

According to the balanced equation,

  $\begin{array}{c}\text{1}\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\to \text{2}\text{mol}\text{of}{\text{O}}_{\text{2}}\\ \text{2}\text{.70}\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\to \text{2}\times \text{2}\text{.70}\text{mol}\text{of}{\text{O}}_{\text{2}}\\ \to 5.40\text{mol}\text{of}{\text{O}}_{\text{2}}\end{array}$

Thus, the oxygen present in excess (6.30 mol is given) so the limiting reagent is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ .

(b)

Interpretation Introduction

Interpretation: The mole of water 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 28SP

The moles of water produced is 5.40 mol.

Explanation of Solution

The limiting reagent is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$. The mol of water is calculated as:

  $\begin{array}{c}\text{1}\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\to \text{2}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \text{2}\text{.70}\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\to \text{2}\times \text{2}\text{.70}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \to 5.40\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Thus, the moles of water produced is 5.40 mol.