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ATOM ECONOMY: Ethylene oxide, C 2 H 4 O, is an important industrial chemical [as it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers]. One way to make the compound is called the “chlorohydrin route.” C 2 H 4 + Cl 2 + Ca(OH) 2 → C 2 H 4 O + CaCl 2 + H 2 O Another route is the modern catalytic reaction. C 2 H 4 + 1/2 O 2 → C 2 H 4 O (a) Calculate the % atom economy for the production of C 2 H 4 O in each of these reactions. Which is the more efficient method? (b) What is the percent yield of C 2 H 4 O if 867 g of C 2 H 4 is used to synthesize 762 g of the product by the catalytic reaction?

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Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

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Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 120GQ
Textbook Problem
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ATOM ECONOMY: Ethylene oxide, C2H4O, is an important industrial chemical [as it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers]. One way to make the compound is called the “chlorohydrin route.”

C2H4 + Cl2 + Ca(OH)2 → C2H4O + CaCl2 + H2O

Another route is the modern catalytic reaction.

C2H4 + 1/2 O2 → C2H4O

  1. (a) Calculate the % atom economy for the production of C2H4O in each of these reactions. Which is the more efficient method?
  2. (b) What is the percent yield of C2H4O if 867 g of C2H4 is used to synthesize 762 g of the product by the catalytic reaction?

(a)

Interpretation Introduction

Interpretation:

The %atomeconomy for the product in the given reactions has to be determined.

Concept introduction:

  • Atom economy is one of the key parameter used to evaluate the efficiency of a reaction.

  %atomeconomy=molarmassofatomutilizedmolarmassofreactants×100%

  • The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
  • Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield and multiplied with hundred.
  • Reactions which are carried out with the help of catalysts are known as catalytic reactions.

Explanation of Solution

Atom economy is one of the key parameters used to evaluate the efficiency of a reaction.

Atom economy percentage of a reaction can be calculated by using the following equation.

  %atomeconomy=molarmassofatomutilizedmolarmassofreactants×100%

In the first reaction preparation of C2H4O has done through the reactions,

  C2H4+Cl2+Ca(OH)2C2H4O+CaCl2+H2O

In the above reaction reactant molecule contains 2C,6H,1O,1Ca,and2Cl so the combined molar mass of the reactants is 157.07g/mol

The product molecule C2H4O contains 2C,4H,and1O and so its molar mass is 44g/mol

     %atomeconomy=molarmassofatomutilizedmolarmassofreactants×100% =(44)(157

(b)

Interpretation Introduction

Interpretation:

Percent yield of given reaction has to be determined.

Concept introduction:

  • Atom economy is one of the key parameter used to evaluate the efficiency of a reaction.

  %atomeconomy=molarmassofatomutilizedmolarmassofreactants×100%

  • The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
  • Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield and multiplied with hundred.
  • Reactions which are carried out with the help of catalysts are known as catalytic reactions.

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Chapter 4 Solutions

Chemistry & Chemical Reactivity
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