Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN: 9781285199023
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Textbook Question
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?

a.

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The balanced chemical equation.

Explanation of Solution

In the combustion the chloroethylene, C2H3Cl reacts with oxygen. The combustion of C2H3Cl

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

C2H3Cl(l) + 52O2(g)2CO2(g)+H2O(g)+HCl(g)

b.

Expert Solution
Check Mark
Interpretation Introduction

To determine:

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

Explanation of Solution

Calculate the molecular weight of C2H3Cl

MW(C2H3Cl)= 2 × 12gmol + 3 × 1 gmol + 1 ×35.5 gmol=62.5 gmol

Calculate the grams of C2H3Cl that are in a mol of this compound. This mol reacts with 2.5 moles of O2 in the reaction.

1 mol C2H3Cl × 62.5 g C2H3Cl1 mol C2H3Cl=62.5 g C2H3Cl

Calculate the moles of O2 that react with 35.00 g of C2H3Cl

62.5 g C2H3Cl-----------2.5 moles O235.00 g C2H3Cl----------- 35 .00 g C2H3Cl x 2 .5 moles O262 .5 g C2H3Cl=1.4 moles O2

c.

Expert Solution
Check Mark
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 CO2, H2O,andHCl

MW(CO2)= 1 ×12gmol + 2 × 16 gmol = 44 gmolMW(H2O)= 2 × 1gmol + 1 × 16 gmol = 18 gmolMW(HCl)= 1 ×1gmol + 1 × 35.5 gmol = 36.5 gmol

Calculate the grams of CO2, H2O,andHCl that react with 1 mol of C2H3Cl in the reaction

2 mol CO2 ×  44 g CO2 1 mol CO2=88 g CO21 mol H2×  18 g H2O 1 mol H2O=18 g H2O1 mol HCl ×36.5 g HCl1 mol HCl=36.5 g HCl

Calculate the grams of CO2, H2O,andHCl that are produced with 25.00 g of C2H3Cl.

62.5 g C2H3Cl-----------88 g CO225.00 g C2H3Cl----------- 25 .00 g C2H3Cl ×  88 g CO262 .5 g C2H3Cl= 35.2 g CO262.5 g C2H3Cl-----------18 g H2O25.00 g C2H3Cl----------- 25 .00 g C2H3Cl ×  18 g H2O62 .5 g C2H3Cl= 7.2 g H2O62.5 g C2H3Cl-----------36.5 g HCl25.00 g C2H3Cl----------- 25 .00 g C2H3Cl × 36.5 g HCl62 .5 g C2H3Cl= 14.6 g HCl

Conclusion

According to the following reaction,

a. C2H3Cl(l) + 52O2(g)2CO2(g)+H2O(g)+HCl(g)

b. 35.00 g of C2H3Cl reacts with 1.4 moles of O2.

c. 25.00 g produces 35.2 g of CO2, 7.2 g of H2O and 14.6 g ofO2.

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Chapter 4 Solutions

Chemistry for Engineering Students

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