Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.155P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction is to be written. Also, the reducing and oxidizing agent is to be identified.

Concept introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to the transfer of an electron from one species to another.

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 4.155P

The balanced equation for the reaction is as follows:

  3NH4ClO4(s)+3Al(aq)SunlightAl2O3(s)+AlCl3(s)+6H2O(g)+3NO(g)

The reducing agent is NH4ClO4 and the oxidizing agent is Al.

Explanation of Solution

The given balanced chemical reaction is as follows:

  3NH4ClO4(s)+3Al(aq)SunlightAl2O3(s)+AlCl3(s)+6H2O(g)+3NO(g)

The oxidation number of aluminium is +3, the oxidation number of hydrogen attached to non-metal is +1 and oxidation number of nitrogen in ammonium ion is 3.

The expression to calculate the oxidation number of chlorine in NH4ClO4 is,

  [(oxidation number of nitrogen)+4(oxidation number of hydrogen)+4(oxidation number of oxygen)+(oxidation number of chloride)]=0        (1)

Rearrange equation (1) for the oxidation number of chlorine.

  Oxidation number of chloride=[(oxidation number of nitrogen)4(oxidation number of hydrogen)4(oxidation number of oxygen)]        (2)

Substitute 2 for the oxidation number of oxygen, 3 for the oxidation number of nitrogen and +1 for the oxidation number of hydrogen in equation (2).

  Oxidation number of chloride=[(3)4(+1)4(2)]=[+34+8]=+7

The expression to calculate the oxidation number of chlorine in AlCl3 is as follows:

  [3(oxidation number of chlorine)+(oxidation number of aluminium)]=0        (3)

Rearrange equation (3) for the oxidation number of chlorine.

  Oxidation number of chlorine=13[(oxidation number of aluminium)]        (4)

Substitute +3 for the oxidation number of aluminium in equation (4).

  Oxidation number of chlorine=13[(+3)]=1

The oxidation state of chlorine in NH4ClO4 is +7 and it decreases to 1 in AlCl3. NH4ClO4 undergoes reduction in the reaction, therefore, it acts as the oxidizing agent.

The oxidation state of aluminium in Al is 0 and it increases to +3 in AlCl3. Al undergoes oxidation in the reaction, therefore, it acts as the reducing agent.

Conclusion

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

(b)

Interpretation Introduction

Interpretation:

The total mole of gas produced when 50.0 kg of ammonium perchlorate reacts with a stoichiometric amount of Al is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

A+2B3C

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

(b)

Expert Solution
Check Mark

Answer to Problem 4.155P

1.28×103mol of gas is produced when 50.0 kg of ammonium perchlorate reacts with a stoichiometric amount of Al.

Explanation of Solution

The given balanced chemical reaction is as follows:

  3NH4ClO4(s)+3Al(aq)SunlightAl2O3(s)+AlCl3(s)+6H2O(g)+3NO(g)

The formula to calculate the moles of NH4ClO4 is as follows:

  Moles of NH4ClO4=[given massof NH4ClO4molecular mass of NH4ClO4]        (5)

Substitute 50.0 kg for the mass of NH4ClO4 and 117.49 g/mol for the molar mass of NH4ClO4 in the equation (5).

  Moles of NH4ClO4=(50.0 kg117.49 g/mol)(1000g1kg)=425.568133mol

The formula to calculate the moles of gas is as follows:

  Moles ofgas=moles ofNH4ClO4(9molgas3molNH4ClO4)        (6)

Substitute 425.568133mol for moles of NH4ClO4 in the equation (6).

  Moles ofgas=(425.568133mol)(9molgas3molNH4ClO4)=1276.70mol1.28×103mol.

Conclusion

Stoichiometry of a reaction can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

(c)

Interpretation Introduction

Interpretation:

The change in volume from the reaction is to be calculated.

Concept introduction:

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is kg/m3. The formula to calculate density is,

  Density=MassVolume        (7)

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  A+2B3C

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The conversion factor to convert mL to cc is as follows:

  1mL=1cc

(c)

Expert Solution
Check Mark

Answer to Problem 4.155P

The change in volume is 2.86×104L.

Explanation of Solution

The given balanced chemical reaction is as follows:

  3NH4ClO4(s)+3Al(aq)SunlightAl2O3(s)+AlCl3(s)+6H2O(g)+3NO(g)

Rearrange the equation (7) to calculate the volume.

  Volume=MassDensity        (8)

Substitute 1.9g/cc for density and 50.0 kg for mass in the equation (8) to calculate the volume of NH4ClO4.

  Volumeof NH4ClO4=(50.0 kg1.9g/cc)(1000g1kg)(1mL1cc)(1L1000mL)=25.6410L

The formula to calculate the mass of Al is as follows:

  Mass ofAl=moles ofNH4ClO4(3molAl3molNH4ClO4)(molar mass of Al)        (9)

Substitute 425.568133mol for moles of NH4ClO4 and 26.98 g/mol for molar mass of Al in the equation (9).

  Mass ofAl=425.568133mol(3molAl3molNH4ClO4)(26.98 g/mol)=11481.828g

Substitute 2.70g/cc for density and 11481.828g for mass in the equation (8) to calculate the volume of Al.

  Volumeof Al=(11481.828g2.70g/cc)(1mL1cc)(1L1000mL)=4.2525L

The formula to calculate the initial volume is as follows:

  Initial Volume=volume of NH4ClO4+volumeofAl        (10)

Substitute 25.6410L for volume of NH4ClO4 and 4.2525L for volume of Al in the equation (10).

  Initial Volume=25.6410L+4.2525L=29.8935 L

The formula to calculate the mass of Al2O3 is as follows:

  Mass ofAl2O3=[moles ofNH4ClO4(1molAl2O33molNH4ClO4)(molar mass of Al2O3)]        (11)

Substitute 425.568133mol for moles of NH4ClO4 and 101.96 g/mol for molar mass of Al2O3 in the equation (11).

  Mass ofAl2O3=[(425.568133mol)(1molAl2O33molNH4ClO4)(101.96 g/mol)]=14463.64g

Substitute 3.97g/cc for density and 14463.64g for mass in the equation (8) to calculate the volume of Al2O3.

Volumeof Al2O3=(14463.64g3.97g/cc)(1mL1cc)(1L1000mL)=3.6432L

The formula to calculate the mass of AlCl3 is as follows:

  Mass ofAlCl3=[moles ofNH4ClO4(1molAlCl33molNH4ClO4)(molar mass of AlCl3)]        (12)

Substitute 425.568133mol for moles of NH4ClO4 and 133.33 g/mol for molar mass of AlCl3 in the equation (12).

  Mass ofAlCl3=[(425.568133mol)(1molAlCl33molNH4ClO4)(133.33 g/mol)]=18913.67g

Substitute 2.44g/cc for density and 18913.67g for mass in the equation (8) to calculate the volume of AlCl3.

  Volumeof AlCl3=(18913.67g2.44g/cc)(1mL1cc)(1L1000mL)=7.7515L

The volume of one mole of gas at STP is 22.4L. The volume of 1276.70mol of gas is calculated as follows:

  Volume ofgas=(1276.70mol)(22.4L1mol)=28598.08L

The formula to calculate the final volume is as follows:

  Final Volume=volume of Al2O3+volumeofAlCl3+volumeofgas        (13)

Substitute 3.6432L for volume of Al2O3, 7.7515L for volume of AlCl3 and 28598.08L for volume of gas in the equation (13).

  Final Volume=3.6432L+7.7515L+28598.08L=28609.4747L

The formula to calculate the volume change is as follows:

  Volume change=Final VolumeInitial Volume        (14)

Substitute 28609.4747L for final volume and 29.8935 L for initial volume in the equation (14).

  Volume change=28609.4747L29.8935 L=28579.5812L2.86×104L.

Conclusion

The volume change in a reaction is defined as the difference between volume of reactants and the volume of products.

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Chapter 4 Solutions

Chemistry: The Molecular Nature of Matter and Change

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