Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781305176461
Author: Kotz
Publisher: Cengage
Question
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Chapter 21, Problem 76GQ

(a)

Interpretation Introduction

Interpretation:

To calculate the mass of H2O and O2 produced in the decomposition of ammonium perchlorate.

Concept introduction:

Ammonium perchlorate upon decomposition gives NO, Cl2 and O2 gases. It is used as the oxidizer in solid-fuel rockets.

The total number of moles of the substance is equal to the mass of the substance divided by the molecular weight of that substance.

The limiting reagent in the chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by the limiting reagent.

(a)

Expert Solution
Check Mark

Answer to Problem 76GQ

The mass of water molecule is produced upon decomposition of ammonium perchlorate is 1.94×105kg.

The mass of oxygen molecule is produced upon decomposition of ammonium perchlorate is 0.864×105kg.

Explanation of Solution

The mass of H2O and O2 produced in the decomposition of ammonium perchlorate is calculated below.

Given:

The mass of ammonium perchlorate is 6.35×105kg.

The balanced chemical equation of decomposition of ammonium perchlorate is written as,

    2NH4ClO4(s)4H2O(l)+O2(g)+Cl2(g)+2NO(g)                          (1)

The number of moles (nNH4ClO4) of ammonium perchlorate is equal to the mass of ammonium perchlorate divided by its molecular weight. The number of moles (nNH4ClO4) is calculated as,

nNH4ClO4=6.35×108g117.49gmol1=0.05×108mol

From equation (1), two moles of ammonium perchlorate produced four moles of water molecules. Therefore, the number of moles (nH2O) of water molecules produced from 0.05×108mol of ammonium perchlorate is written as,

nH2O=(4mol2mol)(0.05×108mol)=0.108×108mol

The mass (mH2O) of water molecule is equal to the product of a number of moles of the water molecule and its molecular weight. The mass (mH2O) of water molecule is calculated as,

mH2O=(0.108×108mol)(18gmol1)=1.94×108g=1.94×105kg

Therefore, the mass of water molecule is produced upon decomposition of ammonium perchlorate is 1.94×105kg.

From equation (1), two moles of ammonium perchlorate produced one mole of the oxygen molecule. Therefore, the number of moles (nO2) of oxygen molecule produced from 0.05×108mol of ammonium perchlorate is written as,

nO2=(1mol2mol)(0.05×108mol)=0.027×108mol

The mass (mO2) of oxygen molecule is equal to the product of a number of moles of the oxygen molecule and its molecular weight. The mass (mO2) of oxygen molecule is calculated as,

mO2=(0.027×108mol)(32gmol1)=0.864×108g=0.864×105kg

Therefore, the mass of oxygen molecule is produced upon decomposition of ammonium perchlorate is 0.864×105kg.

(b)

Interpretation Introduction

Interpretation:

To calculate the mass of aluminium which is required to use up all of the oxygen to form Al2O3

Concept introduction:

Ammonium perchlorate upon decomposition gives NO, Cl2 and O2 gases. It is used as the oxidizer in solid-fuel rockets.

The total number of moles of the substance is equal to the mass of the substance divided by the molecular weight of that substance.

The limiting reagent in the chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by the limiting reagent.

(b)

Expert Solution
Check Mark

Answer to Problem 76GQ

The mass of aluminium powder which is required to use up all of the oxygen to form Al2O3 is 0.97×105kg.

Explanation of Solution

The mass of aluminium which is required to use up all of the oxygen to form Al2O3 is calculated below.

Given:

The number of moles of oxygen molecule calculated in sub-part (a) is 0.027×108mol.

The balanced chemical equation of the reaction of aluminium with oxygen molecule is written as,

    4Al(s)+3O2(g)2Al2O3(s) (2)

Since the entire oxygen molecule is used to react with the aluminium powder to produce Al2O3. Hence, the number of moles of aluminium is equal to the number of moles of the oxygen molecule.

From equation (2), four moles of aluminium reacted with three moles of the oxygen molecule. Therefore, the number of moles (nAl) of aluminium is calculated as,

nAl=(4mol3mol)(0.027×108mol)=0.036×108mol

The mass (mAl) of aluminium powder is equal to the product of a number of moles of aluminium powder and its molecular weight. The mass (mAl) of aluminium powder is calculated as,

mAl=(0.036×108mol)(26.98gmol1)=0.97×108g=0.97×105kg

Therefore, the mass of aluminium powder which is required to use up all of the oxygen to form Al2O3 is 0.97×105kg.

(c)

Interpretation Introduction

Interpretation:

To calculate the mass of Al2O3 produced in the reaction of powdered aluminium with oxygen.

Concept introduction:

Ammonium perchlorate upon decomposition gives NO, Cl2 and O2 gases. It is used as the oxidizer in solid-fuel rockets.

The total number of moles of the substance is equal to the mass of the substance divided by the molecular weight of that substance.

The limiting reagent in the chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by the limiting reagent.

(c)

Expert Solution
Check Mark

Answer to Problem 76GQ

The mass of Al2O3 produced in the reaction of powdered aluminium with oxygen is 1.83×105kg.

Explanation of Solution

The mass of Al2O3 produced in the reaction of powdered aluminium with oxygen is calculated below.

Given:

The mass of aluminium powder which is required to use up all of the oxygen to form Al2O3 calculated in sub-part (b) is 0.97×105kg.

The balanced chemical equation of the reaction of aluminium with oxygen molecule is written as,

    4Al(s)+3O2(g)2Al2O3(s)

Four moles of aluminium produced two moles of Al2O3 molecule. Therefore, the number of moles (nAl2O3) of Al2O3 molecule is calculated as,

nAl2O3=(2mol4mol)(0.036×108mol)=0.018×108mol

The mass (mAl2O3) of Al2O3 molecule is equal to the product of a number of moles of Al2O3 molecule and its molecular weight. The mass (mAl2O3) of Al2O3 molecule is calculated as,

mAl=(0.018×108mol)(101.96gmol1)=1.83×108g=1.83×105kg

Therefore, the mass of Al2O3 produced in the reaction of powdered aluminium with oxygen is 1.83×105kg.

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Chapter 21 Solutions

Chemistry & Chemical Reactivity

Ch. 21.8 - Prob. 4QCh. 21.8 - Prob. 3RCCh. 21.11 - Prob. 1QCh. 21.11 - Prob. 2QCh. 21 - Give examples of two basic oxides. Write equations...Ch. 21 - Prob. 2PSCh. 21 - Prob. 3PSCh. 21 - Prob. 4PSCh. 21 - Prob. 5PSCh. 21 - Prob. 6PSCh. 21 - For the product of the reaction you selected in...Ch. 21 - For the product of the reaction you selected in...Ch. 21 - Prob. 9PSCh. 21 - Prob. 10PSCh. 21 - Place the following oxides in order of increasing...Ch. 21 - Place the following oxides in order of increasing...Ch. 21 - Prob. 13PSCh. 21 - Prob. 14PSCh. 21 - Prob. 15PSCh. 21 - Prob. 16PSCh. 21 - Prob. 17PSCh. 21 - Prob. 18PSCh. 21 - Prob. 19PSCh. 21 - Prob. 20PSCh. 21 - Prob. 21PSCh. 21 - Write balanced equations for the reaction of...Ch. 21 - Prob. 23PSCh. 21 - (a) Write equations for the half-reactions that...Ch. 21 - When magnesium bums in air, it forms both an oxide...Ch. 21 - Prob. 26PSCh. 21 - Prob. 27PSCh. 21 - Prob. 28PSCh. 21 - Calcium oxide, CaO, is used to remove SO2 from...Ch. 21 - Prob. 30PSCh. 21 - Prob. 31PSCh. 21 - The boron trihalides (except BF3) hydrolyze...Ch. 21 - When boron hydrides burn in air, the reactions are...Ch. 21 - Prob. 34PSCh. 21 - Write balanced equations for the reactions of...Ch. 21 - Prob. 36PSCh. 21 - Prob. 37PSCh. 21 - Alumina, Al2O3, is amphoteric. Among examples of...Ch. 21 - Prob. 39PSCh. 21 - Prob. 40PSCh. 21 - Describe the structure of pyroxenes (see page...Ch. 21 - Describe how ultrapure silicon can be produced...Ch. 21 - Prob. 43PSCh. 21 - Prob. 44PSCh. 21 - Prob. 45PSCh. 21 - Prob. 46PSCh. 21 - Prob. 47PSCh. 21 - The overall reaction involved in the industrial...Ch. 21 - Prob. 49PSCh. 21 - Prob. 50PSCh. 21 - Prob. 51PSCh. 21 - Prob. 52PSCh. 21 - Prob. 53PSCh. 21 - Prob. 54PSCh. 21 - Prob. 55PSCh. 21 - Sulfur forms a range of compounds with fluorine....Ch. 21 - The halogen oxides and oxoanions are good...Ch. 21 - Prob. 58PSCh. 21 - Bromine is obtained from brine wells. The process...Ch. 21 - Prob. 60PSCh. 21 - Prob. 61PSCh. 21 - Halogens combine with one another to produce...Ch. 21 - The standard enthalpy of formation of XeF4 is 218...Ch. 21 - Draw the Lewis electron dot structure for XeO3F2....Ch. 21 - Prob. 65PSCh. 21 - Prob. 66PSCh. 21 - Prob. 67GQCh. 21 - Prob. 68GQCh. 21 - Consider the chemistries of the elements...Ch. 21 - When BCl3 gas is passed through an electric...Ch. 21 - Prob. 71GQCh. 21 - Prob. 72GQCh. 21 - Prob. 73GQCh. 21 - Prob. 74GQCh. 21 - Prob. 75GQCh. 21 - Prob. 76GQCh. 21 - Prob. 77GQCh. 21 - Prob. 78GQCh. 21 - Prob. 79GQCh. 21 - Prob. 80GQCh. 21 - Prob. 81GQCh. 21 - Prob. 83GQCh. 21 - Prob. 84GQCh. 21 - A Boron and hydrogen form an extensive family of...Ch. 21 - In 1774, C. Scheele obtained a gas by reacting...Ch. 21 - What current must be used in a Downs cell...Ch. 21 - The chemistry of gallium: (a) Gallium hydroxide,...Ch. 21 - Prob. 89GQCh. 21 - Prob. 90GQCh. 21 - Prob. 91GQCh. 21 - Prob. 92GQCh. 21 - Prob. 93ILCh. 21 - Prob. 94ILCh. 21 - Prob. 95ILCh. 21 - Prob. 96ILCh. 21 - Prob. 97ILCh. 21 - Prob. 98ILCh. 21 - Prob. 99SCQCh. 21 - Prob. 100SCQCh. 21 - Prob. 101SCQCh. 21 - Prob. 102SCQCh. 21 - Prob. 103SCQCh. 21 - Prob. 104SCQCh. 21 - Prob. 105SCQCh. 21 - Prob. 106SCQCh. 21 - Prob. 107SCQCh. 21 - Prob. 108SCQCh. 21 - Prob. 109SCQCh. 21 - Prob. 110SCQCh. 21 - Comparing the chemistry of carbon and silicon. (a)...Ch. 21 - Prob. 112SCQCh. 21 - Xenon trioxide, XeO3, reacts with aqueous base to...
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