Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 12, Problem 46A

a.

Interpretation Introduction

Interpretation: To calculate the moles of each reactant.

Concept Introduction: A mole is 6.02×1023 atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

a.

Expert Solution
Check Mark

Answer to Problem 46A

The number of moles of carbon monoxide is 11.3 moles.

The number of moles of hydrogen is 22.5 moles.

Explanation of Solution

The given chemical reaction is depicted as follows:

  COg+2H2gCH3OHg

When one mole of carbon monoxide reacts with two moles of hydrogen, it yields one mole of methanol.

The grams of methanol are given i.e., 3.60×102 g .

The molar mass of methanol is 32.0 g mol1 .

Therefore, the moles of carbon monoxide can be calculated as follows:

  3.60×102g CH3OH×1 mol CH3OH32.0 g CH3OH×1 mol CO1 mol CH3OH=11.3 mol CO

Thus, the number of moles of carbon monoxide is 11.3 moles.

The calculation for moles of hydrogen is as follows:

  3.60×102g CH3OH×1 mol CH3OH32.0 g CH3OH×2 mol H21 mol CH3OH=22.5 mol H2

Thus, the number of moles of hydrogen is 22.5 moles.

b.

Interpretation Introduction

Interpretation: To calculate the number of grams of each reactant.

Concept Introduction: A mole is 6.02×1023 atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. In a balanced chemical equation, the coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

b.

Expert Solution
Check Mark

Answer to Problem 46A

The grams of carbon monoxide are 112 grams.

The grams of hydrogen are 16 grams.

Explanation of Solution

The given chemical reaction is depicted as follows:

  COg+2H2gCH3OHg

When one mole of carbon monoxide reacts with two moles of hydrogen, it yields one mole of methanol.

The moles of methanol are given i.e., 4.00 mol .

The molar mass of carbon monoxide is 28.0 g mol1 .

Therefore, the grams of carbon monoxide can be calculated as follows:

  4.00 mol CH3OH×1 mol CO1mol CH3OH×28.0 g CO1 mol CO=112 g CO

Thus, the grams of carbon monoxide are 112 grams.

The calculation for grams of hydrogen is as follows:

The molar mass of hydrogen is 2.0 g mol1 .

  4.00 mol CH3OH×mol H21mol CH3OH×2.0 g H21 mol H2=16 g H2

Thus, the grams of hydrogen are 16 grams.

c.

Interpretation Introduction

Interpretation: To calculate the grams of hydrogen needed to react.

Concept Introduction: A mole is 6.02×1023 atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. In a balanced chemical equation, the coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

c.

Expert Solution
Check Mark

Answer to Problem 46A

The grams of hydrogen are 11 grams.

Explanation of Solution

The given chemical reaction is depicted as follows:

  COg+2H2gCH3OHg

When one mole of carbon monoxide reacts with two moles of hydrogen, it yields one mole of methanol.

The moles of carbon monoxide is given, i.e., 2.85 mol .

The calculation for grams of hydrogen is as follows:

The molar mass of hydrogen is 2.0 g mol1 .

  2.85mol CO×mol H21mol CO×2.0 g H21 mol H2=11 g H2

Thus, the grams of hydrogen are 11 grams.

Chapter 12 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 12.2 - Prob. 12SPCh. 12.2 - Prob. 13SPCh. 12.2 - Prob. 14SPCh. 12.2 - Prob. 15SPCh. 12.2 - Prob. 16SPCh. 12.2 - Prob. 17SPCh. 12.2 - Prob. 18SPCh. 12.2 - Prob. 19SPCh. 12.2 - Prob. 20SPCh. 12.2 - Prob. 21LCCh. 12.2 - Prob. 22LCCh. 12.2 - Prob. 23LCCh. 12.2 - Prob. 24LCCh. 12.2 - Prob. 25LCCh. 12.3 - Prob. 26SPCh. 12.3 - Prob. 27SPCh. 12.3 - Prob. 28SPCh. 12.3 - Prob. 29SPCh. 12.3 - Prob. 30SPCh. 12.3 - Prob. 31SPCh. 12.3 - Prob. 32SPCh. 12.3 - Prob. 33SPCh. 12.3 - Prob. 34LCCh. 12.3 - Prob. 35LCCh. 12.3 - Prob. 36LCCh. 12.3 - Prob. 37LCCh. 12.3 - Prob. 38LCCh. 12 - Prob. 39ACh. 12 - Prob. 40ACh. 12 - Prob. 41ACh. 12 - Prob. 42ACh. 12 - Prob. 43ACh. 12 - Prob. 44ACh. 12 - Prob. 45ACh. 12 - Prob. 46ACh. 12 - Prob. 47ACh. 12 - Prob. 48ACh. 12 - Prob. 49ACh. 12 - Prob. 50ACh. 12 - Prob. 51ACh. 12 - Prob. 52ACh. 12 - Prob. 53ACh. 12 - Prob. 54ACh. 12 - Prob. 55ACh. 12 - Prob. 56ACh. 12 - Prob. 57ACh. 12 - Prob. 58ACh. 12 - Prob. 59ACh. 12 - Prob. 60ACh. 12 - Prob. 61ACh. 12 - Prob. 62ACh. 12 - Prob. 63ACh. 12 - Prob. 64ACh. 12 - Prob. 65ACh. 12 - Prob. 66ACh. 12 - Prob. 67ACh. 12 - Prob. 68ACh. 12 - Prob. 69ACh. 12 - Prob. 70ACh. 12 - Prob. 71ACh. 12 - Prob. 72ACh. 12 - Prob. 73ACh. 12 - Prob. 74ACh. 12 - Prob. 75ACh. 12 - Prob. 76ACh. 12 - Prob. 77ACh. 12 - Prob. 78ACh. 12 - Prob. 79ACh. 12 - Prob. 80ACh. 12 - Prob. 81ACh. 12 - Prob. 82ACh. 12 - Prob. 83ACh. 12 - Prob. 84ACh. 12 - Prob. 85ACh. 12 - Prob. 88ACh. 12 - Prob. 89ACh. 12 - Prob. 90ACh. 12 - Prob. 91ACh. 12 - Prob. 92ACh. 12 - Prob. 93ACh. 12 - Prob. 94ACh. 12 - Prob. 95ACh. 12 - Prob. 96ACh. 12 - Prob. 97ACh. 12 - Prob. 98ACh. 12 - Prob. 99ACh. 12 - Prob. 100ACh. 12 - Prob. 101ACh. 12 - Prob. 102ACh. 12 - Prob. 103ACh. 12 - Prob. 104ACh. 12 - Prob. 105ACh. 12 - Prob. 106ACh. 12 - Prob. 107ACh. 12 - Prob. 1STPCh. 12 - Prob. 2STPCh. 12 - Prob. 3STPCh. 12 - Prob. 4STPCh. 12 - Prob. 5STPCh. 12 - Prob. 6STPCh. 12 - Prob. 7STPCh. 12 - Prob. 8STPCh. 12 - Prob. 9STPCh. 12 - Prob. 10STP
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