Chapter 12.3, Problem 29SP

Interpretation Introduction

Interpretation: The grams 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 29SP

The grams of water is 43.2 g.

Explanation of Solution

Given:

• 2.40 mol ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$
• 7.40 mol ${\text{O}}_{\text{2}}$

The reaction is given as follows:

  $2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to {\text{4CO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

According to the balanced equation,

  $\begin{array}{c}2\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 5\text{mol}\text{of}{\text{O}}_{\text{2}}\\ 1\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to \frac{5}{2}\text{mol}\text{of}{\text{O}}_{\text{2}}\\ 2.40\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to \frac{5}{2}\times 2.40\text{mol}\text{of}{\text{O}}_{\text{2}}\\ \to 6.0\text{mol}\text{of}{\text{O}}_{\text{2}}\end{array}$

The mol of ${\text{O}}_{\text{2}}$ is given as 7.40 mol means that it is present in excess. So, the limiting reagent is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$. The mol of water is calculated as follows:

  $\begin{array}{c}2\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 2\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}\\ 1\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 1\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}\\ 2.40\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 2.40\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

The mass of water is calculated as:

  $\begin{array}{c}\text{Mass}=\text{Mole}\times \text{Molar}\text{mass}\\ =2.40\text{mol}\times 18\text{g/mol}\\ =\text{43}\text{.2}\text{g}\end{array}$

Thus, the grams of water is 43.2 g.