Chapter 4, Problem 4.14PAE

4.14 The combustion of liquid chloroethylene, C₂H₃Cl, yields carbon dioxide, steam, and hydrogen chloride gas. (a) Write a balanced equation for the reaction. (b) How many moles of oxygen are required to react with 35.00 g of chloroethylene? (c) If 125.00 g of chloroethylene reacts with an excess of oxygen, how many grams of each product are formed?

Interpretation Introduction

To determine:

The balanced chemical equation.

Explanation of Solution

In the combustion the chloroethylene, ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ reacts with oxygen. The combustion of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$

produces carbon dioxide, steam, and hydrogen chloride gas, so that, the balanced equation is:

${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl(l) + }\frac{5}{2}{\text{O}}_2(g)\to 2{\text{CO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(g)+HCl(g)}$

Interpretation Introduction

To determine:

Moles of oxygen required to react with 35.00 g of chloroethylene.

Explanation of Solution

Calculate the molecular weight of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$

${\text{MW(C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl)= 2 }\times \text{ 12}\frac{\text{g}}{\text{mol}}\text{ + 3 }\times \text{ 1 }\frac{\text{g}}{\text{mol}}\text{ + 1 }\times \text{35}\text{.5 }\frac{\text{g}}{\text{mol}}=\text{62}\text{.5 }\frac{\text{g}}{\text{mol}}$

Calculate the grams of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ that are in a mol of this compound. This mol reacts with 2.5 moles of ${\text{O}}_{\text{2}}$ in the reaction.

$1{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl }\times \frac{\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}}=\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$

Calculate the moles of ${\text{O}}_{\text{2}}$ that react with 35.00 g of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$

$\begin{array}{l}\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl-----------2}{\text{.5 moles O}}_2\\ \text{35}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl----------- }\frac{\text{35}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl x 2}{\text{.5 moles O}}_2}{\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}}\text{=1}{\text{.4 moles O}}_2\end{array}$

Interpretation Introduction

To determine:Grams of each product formed if 25.00 g of chloroethylene reacts with excess of oxygen.

Explanation of Solution

Calculate the molecular weight of ${\text{CO}}_{\text{2}},{\text{ H}}_{\text{2}}\text{O},\text{and}\text{HCl}$

$\begin{array}{l}{\text{MW(CO}}_{\text{2}}\text{)= 1 }\times \text{12}\frac{\text{g}}{\text{mol}}\text{ + 2 }\times \text{ 16 }\frac{\text{g}}{\text{mol}}=44\frac{\text{g}}{\text{mol}}\\ {\text{MW(H}}_{\text{2}}\text{O)= 2 }\times \text{ 1}\frac{\text{g}}{\text{mol}}\text{ + 1 }\times \text{ 16 }\frac{\text{g}}{\text{mol}}=\text{ 18 }\frac{\text{g}}{\text{mol}}\\ \text{MW(HCl)= 1 }\times \text{1}\frac{\text{g}}{\text{mol}}\text{ + 1 }\times \text{ 35}\text{.5 }\frac{\text{g}}{\text{mol}}=\text{ 36}\text{.5 }\frac{\text{g}}{\text{mol}}\end{array}$

Calculate the grams of ${\text{CO}}_{\text{2}},{\text{ H}}_{\text{2}}\text{O},\text{and}\text{HCl}$ that react with 1 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ in the reaction

$\begin{array}{l}{\text{2 mol CO}}_{\text{2}}\times \frac{{\text{44 g CO}}_{\text{2}}}{{\text{1 mol CO}}_{\text{2}}}=88{\text{ g CO}}_{\text{2}}\\ {\text{1 mol H}}_{\text{2}}\text{O }\times \frac{{\text{18 g H}}_{\text{2}}\text{O}}{{\text{1 mol H}}_{\text{2}}\text{O}}{\text{=18 g H}}_{\text{2}}\text{O}\\ \text{1 mol HCl }\times \frac{\text{36}\text{.5 g HCl}}{\text{1 mol HCl}}=36.5\text{ g HCl}\end{array}$

Calculate the grams of ${\text{CO}}_{\text{2}},{\text{ H}}_{\text{2}}\text{O},\text{and}\text{HCl}$ that are produced with 25.00 g of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$.

$\begin{array}{l}\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{Cl-----------88 g CO}}_2\\ \text{25}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl----------- }\frac{\text{25}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl }\times {\text{ 88 g CO}}_2}{\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}}\text{= 35}{\text{.2 g CO}}_2\\ \text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{Cl-----------18 g H}}_{\text{2}}\text{O}\\ \text{25}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl----------- }\frac{\text{25}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl }\times {\text{ 18 g H}}_{\text{2}}\text{O}}{\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}}\text{= 7}{\text{.2 g H}}_{\text{2}}\text{O}\\ \text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl-----------36}\text{.5 g HCl}\\ \text{25}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl----------- }\frac{\text{25}{\text{.00 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl }\times \text{ 36}\text{.5 g HCl}}{\text{62}{\text{.5 g C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}}\text{= 14}\text{.6 g HCl}\end{array}$

Conclusion

According to the following reaction,

a. ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl(l) + }\frac{5}{2}{\text{O}}_2(g)\to 2{\text{CO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(g)+HCl(g)}$

b. 35.00 g of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ reacts with 1.4 moles of ${\text{O}}_{\text{2}}$.

c. 25.00 g produces $35.{\text{2 g of CO}}_{\text{2}},\text{ 7}.{\text{2 g of H}}_{\text{2}}\text{O and 14}.\text{6 g of}{\text{O}}_{\text{2}}$.