EBK INTRODUCTION TO CHEMISTRY
EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 6, Problem 25QP

(a)

Interpretation Introduction

Interpretation:

The given equation Crs+Cl2gCrCl3s is to be balanced and after that, the number of grams of the second reactant that is required to react completely with 0.600 g of the first reactant is to be determined.

(a)

Expert Solution
Check Mark

Explanation of Solution

The mass of the first reactant is 0.600 g .

The relation among the number of moles, molar mass and mass is,

n= mMM

Here, n is the number of moles, m is mass and MM is molar mass.

By rearranging the above equation mass can be determined as,

m=n×MM …..(1)

The balanced equation is,

2Crs+3Cl2g2CrCl3s

The molar mass of Cr is 52.00g/mol and Cl2 is 70.90g/mol . The mass of Cl2 is determined by using equation (1) as follows:

m=0.600g Cr×mol Cr52.00g Cr×3 mol Cl22mol Cr×70.90g Cl21mol Cl2=1.23 g Cl2

(b)

Interpretation Introduction

Interpretation:

The given equation RbO2s+H2OlO2g+RbOHs is to be balanced and after that, the number of grams of the second reactant that is required to react completely with 0.600 g of the first reactant is to be determined.

(b)

Expert Solution
Check Mark

Explanation of Solution

The balanced equation is,

4RbO2s+2H2Ol3O2g+4RbOHs

The molar mass of RbO2 is 117.47g/mol and H2O is 18.02g/mol . The mass of H2O is determined by using equation (1) as follows:

m=0.600g RbO2×mol RbO2117.47g RbO2×2 mol H2O4mol RbO2×18.02g H2O1mol H2O=0.0460 g H2O

(c)

Interpretation Introduction

Interpretation:

The given equation C5H12g+O2gCO2g+H2Og is to be balanced and after that, the number of grams of the second reactant that is required to react completely with 0.600 g of the first reactant is to be determined.

(c)

Expert Solution
Check Mark

Explanation of Solution

The balanced equation is,

C5H12g+8O2g5CO2g+6H2Og

The molar mass of C5H12 is 72.15g/mol and O2 is 32.00g/mol . The mass of O2 is determined by using equation (1) as follows:

m=0.600g C5H12×mol C5H1272.15g C5H12×8 mol O21mol C5H12×32.00g O21mol O2=2.13 g O2

(d)

Interpretation Introduction

Interpretation:

The given equation Lis+Cl2gLiCls is to be balanced and after that, the number of grams of the second reactant that is required to react completely with 0.600 g of the first reactant is to be determined.

(d)

Expert Solution
Check Mark

Explanation of Solution

The balanced equation is,

2Lis+Cl2g2LiCls

The molar mass of Li is 6.94g/mol and Cl2 is 70.90g/mol . The mass of O2 is determined by using equation (1) as follows:

m=0.600g Li×mol Li6.94g Li×1 mol Cl22mol Li×70.90g Cl21mol Cl2=3.06 g Cl2

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Chapter 6 Solutions

EBK INTRODUCTION TO CHEMISTRY

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