   Chapter 3, Problem 131E

Chapter
Section
Textbook Problem

The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCI). In addition, the reaction invariably produces a variety of other minor products, including C2H4Cl2, C2H3Cl3, and others. Naturally, the production of these minor products reduces the yield of the main product. Calculate the percent yield of C2H5Cl if the reaction of 300. g of ethane with 650. g of chlorine produced 490. g of C2H5Cl

Interpretation Introduction

Interpretation: The reaction between ethane gas and chlorine is given. The percent yield of C2H5Cl is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

Massofthesubstance=(Numberofmoles)×(Molarmassofthesubstance)

To determine: The percent yield of C2H5Cl .

Explanation

Given

The balanced chemical equation for the given chemical reaction is,

C2H6(g)+Cl2(g)C2H5Cl(g)+HCl(g)

The given mass of C2H6 is 300g .

The given mass of Cl2 is 650g .

The given mass of C2H5Cl is 490g .

The molar mass of C2H6 =2C+6H=((2×12)+(6×1))g/mol=30g/mol

The molar mass of Cl2 =2Cl=(2×35.5)g/mol=71g/mol

The molar mass of C2H5Cl =2C+5H+Cl=((2×12)+(5×1)+35.5)g/mol=64.5g/mol

Formula

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassofthesubstanceMolarmassofthesubstance

Substitute the value of the given mass and the molar mass of C2H6 and Cl2 in the above expression.

NumberofmolesofC2H6=300g30g/mol=10mol

NumberofmolesofCl2=650g71g/mol=9

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