CHEMICAL PRINCIPLES PKG W/SAPLING
CHEMICAL PRINCIPLES PKG W/SAPLING
7th Edition
ISBN: 9781319086411
Author: ATKINS
Publisher: MAC HIGHER
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Chapter F, Problem H.5E

(a)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    NaBH4(s)+H2O(l)NaBO2(aq)+H2(g)

Concept Introduction:

Chemical equation is a short form representation of a chemical reaction with all the required conditions necessary for the reaction to take place.  As the atoms are neither created nor destroyed in a chemical reaction, the number of atoms of same element has to be equal on both sides of the equation.  This is known as balanced chemical equation.  Coefficients can be changed but not the subscript.

(a)

Expert Solution
Check Mark

Answer to Problem H.5E

Balanced chemical equation is NaBH4(s)+2H2O(l)NaBO2(aq)+4H2(g).

Explanation of Solution

Given skeletal equation is shown below;

    NaBH4(s)+H2O(l)NaBO2(aq)+H2(g)

Balancing oxygen atoms:  In the reactant side, there is one oxygen atom while on the product side, there are two oxygen atoms.  Adding coefficient 2 before H2O balances the oxygen atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    NaBH4(s)+2H2O(l)NaBO2(aq)+H2(g)

Balancing hydrogen atoms:  In the reactant side, there are eight hydrogen atoms while on the product side there are only two hydrogen atoms.  Adding coefficient 4 before H2 on the product side balances the hydrogen atoms on both sides of the equation.  This step balances the other atoms also.  Therefore, the balanced chemical equation can be written as follows;

    NaBH4(s)+2H2O(l)NaBO2(aq)+4H2(g)

(b)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    Mg(N3)2(s)+H2O(l)Mg(OH)2(aq)+HN3(aq)

Concept Introduction:

Refer part (a).

(b)

Expert Solution
Check Mark

Answer to Problem H.5E

Balanced chemical equation is Mg(N3)2(s)+2H2O(l)Mg(OH)2(aq)+2HN3(aq).

Explanation of Solution

Given skeletal equation is shown below;

    Mg(N3)2(s)+H2O(l)Mg(OH)2(aq)+HN3(aq)

Balancing oxygen atoms:  In the reactant side, there is one oxygen atom while on the product side, there are two oxygen atoms.  Adding coefficient 2 before H2O balances the oxygen atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    Mg(N3)2(s)+2H2O(l)Mg(OH)2(aq)+HN3(aq)

Balancing hydrogen atoms:  In the reactant side, there are four hydrogen atoms while on the product side there are only three hydrogen atoms.  Adding coefficient 2 before HN3 on the product side balances the hydrogen atoms on both sides of the equation.  This step balances the other atoms also.  Therefore, the balanced chemical equation can be written as follows;

    Mg(N3)2(s)+2H2O(l)Mg(OH)2(aq)+2HN3(aq)

(c)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    NaCl(aq)+SO3(g)+H2O(l)Na2SO4(aq)+HCl(aq)

Concept Introduction:

Refer part (a).

(c)

Expert Solution
Check Mark

Answer to Problem H.5E

Balanced chemical equation is 2NaCl(aq)+SO3(g)+H2O(l)Na2SO4(aq)+2HCl(aq).

Explanation of Solution

Given skeletal equation is shown below;

    NaCl(aq)+SO3(g)+H2O(l)Na2SO4(aq)+HCl(aq)

Balancing sodium atoms:  In the reactant side, there is one sodium atom while on the product side, there are two sodium atoms.  Adding coefficient 2 before NaCl balances the sodium atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    2NaCl(aq)+SO3(g)+H2O(l)Na2SO4(aq)+HCl(aq)

Balancing chlorine atoms:  In the reactant side, there are two chlorine atoms while on the product side there is only one chlorine atom.  Adding coefficient 2 before HCl on the product side balances the chlorine atoms on both sides of the equation.  This step balances the other atoms also.  Therefore, the balanced chemical equation can be written as follows;

    2NaCl(aq)+SO3(g)+H2O(l)Na2SO4(aq)+2HCl(aq)

(d)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    Fe2P(s)+S(s)P4S10(s)+FeS(s)

Concept Introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Answer to Problem H.5E

Balanced chemical equation is 4Fe2P(s)+18S(s)P4S10(s)+8FeS(s).

Explanation of Solution

Given skeletal equation is shown below;

    Fe2P(s)+S(s)P4S10(s)+FeS(s)

Balancing phosphorus atoms:  In the reactant side, there is one phosphorus atom while on the product side, there are four phosphorus atoms.  Adding coefficient 4 before Fe2P balances the phosphorus atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    4Fe2P(s)+S(s)P4S10(s)+FeS(s)

Balancing iron atoms:  In the reactant side, there are eight iron atoms while on the product side there is only one iron atom.  Adding coefficient 8 before FeS on the product side balances the iron atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    4Fe2P(s)+S(s)P4S10(s)+8FeS(s)

Balancing sulfur atoms:  In the reactant side, there is only one sulfur atom while on the product side there are eighteen atom.  Adding coefficient 18 before S on the reactant side balances the sulfur atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    4Fe2P(s)+18S(s)P4S10(s)+8FeS(s)

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Chapter F Solutions

CHEMICAL PRINCIPLES PKG W/SAPLING

Ch. F - Prob. A.1ECh. F - Prob. A.2ECh. F - Prob. A.3ECh. F - Prob. A.4ECh. F - Prob. A.5ECh. F - Prob. A.6ECh. F - Prob. A.7ECh. F - Prob. A.8ECh. F - Prob. A.9ECh. F - Prob. A.10ECh. F - Prob. A.11ECh. F - Prob. A.12ECh. F - Prob. A.13ECh. F - Prob. A.14ECh. F - Prob. A.15ECh. F - Prob. A.16ECh. F - Prob. A.17ECh. F - Prob. A.18ECh. F - Prob. A.19ECh. F - Prob. A.20ECh. F - Prob. A.21ECh. F - Prob. A.22ECh. F - Prob. A.23ECh. F - Prob. A.24ECh. F - Prob. A.25ECh. F - Prob. A.26ECh. F - Prob. A.27ECh. F - Prob. A.28ECh. F - Prob. A.29ECh. F - Prob. A.30ECh. F - Prob. A.31ECh. F - Prob. A.32ECh. F - Prob. A.33ECh. F - Prob. A.34ECh. F - Prob. A.35ECh. F - Prob. A.36ECh. F - Prob. A.37ECh. F - Prob. A.38ECh. F - Prob. A.39ECh. F - Prob. A.40ECh. F - Prob. A.41ECh. F - Prob. A.42ECh. F - Prob. B.1ASTCh. F - Prob. B.1BSTCh. F - Prob. B.2ASTCh. F - Prob. B.2BSTCh. F - Prob. B.3ASTCh. F - Prob. B.3BSTCh. F - Prob. B.1ECh. F - Prob. B.2ECh. F - Prob. B.3ECh. F - Prob. B.4ECh. F - Prob. 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F - Prob. D.5BSTCh. F - Prob. D.1ECh. F - Prob. D.2ECh. F - Prob. D.3ECh. F - Prob. D.4ECh. F - Prob. D.5ECh. F - Prob. D.6ECh. F - Prob. D.7ECh. F - Prob. D.8ECh. F - Prob. D.9ECh. F - Prob. D.10ECh. F - Prob. D.11ECh. F - Prob. D.12ECh. F - Prob. D.13ECh. F - Prob. D.14ECh. F - Prob. D.15ECh. F - Prob. D.16ECh. F - Prob. D.17ECh. F - Prob. D.18ECh. F - Prob. D.19ECh. F - Prob. D.20ECh. F - Prob. D.21ECh. F - Prob. D.22ECh. F - Prob. D.23ECh. F - Prob. D.24ECh. F - Prob. D.25ECh. F - Prob. D.26ECh. F - Prob. D.27ECh. F - Prob. D.28ECh. F - Prob. D.29ECh. F - Prob. D.30ECh. F - Prob. D.31ECh. F - Prob. D.32ECh. F - Prob. D.33ECh. F - Prob. D.34ECh. F - Prob. D.35ECh. F - Prob. D.36ECh. F - Prob. E.1ASTCh. F - Prob. E.1BSTCh. F - Prob. E.2ASTCh. F - Prob. E.2BSTCh. F - Prob. E.3ASTCh. F - Prob. E.3BSTCh. F - Prob. E.4ASTCh. F - Prob. E.4BSTCh. F - Prob. E.5ASTCh. F - Prob. E.5BSTCh. F - Prob. E.6ASTCh. F - Prob. E.6BSTCh. F - Prob. E.1ECh. F - Prob. E.2ECh. F - Prob. E.3ECh. F - Prob. 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F - Prob. H.2ECh. F - Prob. H.3ECh. F - Prob. H.4ECh. F - Prob. H.5ECh. F - Prob. H.6ECh. F - Prob. H.7ECh. F - Prob. H.8ECh. F - Prob. H.9ECh. F - Prob. H.10ECh. F - Prob. H.11ECh. F - Prob. H.12ECh. F - Prob. H.13ECh. F - Prob. H.14ECh. F - Prob. H.15ECh. F - Prob. H.16ECh. F - Prob. H.17ECh. F - Prob. H.18ECh. F - Prob. H.19ECh. F - Prob. H.20ECh. F - Prob. H.21ECh. F - Prob. H.22ECh. F - Prob. H.23ECh. F - Prob. H.24ECh. F - Prob. H.25ECh. F - Prob. H.26ECh. F - Prob. I.1ASTCh. F - Prob. I.1BSTCh. F - Prob. I.2ASTCh. F - Prob. I.2BSTCh. F - Prob. I.3ASTCh. F - Prob. I.3BSTCh. F - Prob. I.1ECh. F - Prob. I.2ECh. F - Prob. I.3ECh. F - Prob. I.4ECh. F - Prob. I.5ECh. F - Prob. I.6ECh. F - Prob. I.7ECh. F - Prob. I.8ECh. F - Prob. I.9ECh. F - Prob. I.10ECh. F - Prob. I.11ECh. F - Prob. I.12ECh. F - Prob. I.13ECh. F - Prob. I.14ECh. F - Prob. I.15ECh. F - Prob. I.16ECh. F - Prob. I.17ECh. F - Prob. I.18ECh. F - Prob. I.19ECh. F - Prob. I.20ECh. F - Prob. I.21ECh. F - Prob. I.22ECh. F - Prob. I.23ECh. F - Prob. I.24ECh. F - Prob. I.25ECh. F - Prob. I.26ECh. F - Prob. J.1ASTCh. F - Prob. J.1BSTCh. F - Prob. J.2ASTCh. F - Prob. J.2BSTCh. F - Prob. J.1ECh. F - Prob. J.2ECh. F - Prob. J.3ECh. F - Prob. J.4ECh. F - Prob. J.5ECh. F - Prob. J.6ECh. F - Prob. J.7ECh. F - Prob. J.8ECh. F - Prob. J.9ECh. F - Prob. J.10ECh. F - Prob. J.11ECh. F - Prob. J.12ECh. F - Prob. J.13ECh. F - Prob. J.14ECh. F - Prob. J.15ECh. F - Prob. J.16ECh. F - Prob. J.17ECh. F - Prob. J.18ECh. F - Prob. J.19ECh. F - Prob. J.20ECh. F - Prob. J.21ECh. F - Prob. J.22ECh. F - Prob. J.23ECh. F - Prob. J.24ECh. F - Prob. K.1ASTCh. F - Prob. K.1BSTCh. F - Prob. K.2ASTCh. F - Prob. K.2BSTCh. F - Prob. K.3ASTCh. F - Prob. K.3BSTCh. F - Prob. K.4ASTCh. F - Prob. K.4BSTCh. F - Prob. K.5ASTCh. F - Prob. K.5BSTCh. F - Prob. K.1ECh. F - Prob. K.2ECh. F - Prob. K.3ECh. F - Prob. K.4ECh. F - Prob. K.5ECh. F - Prob. K.6ECh. F - Prob. K.7ECh. F - Prob. K.8ECh. F - Prob. K.9ECh. F - Prob. K.10ECh. F - Prob. K.11ECh. F - Prob. K.12ECh. F - Prob. K.13ECh. F - Prob. K.14ECh. F - Prob. K.15ECh. F - Prob. K.16ECh. F - Prob. K.17ECh. F - Prob. K.18ECh. F - Prob. K.19ECh. F - Prob. K.20ECh. F - Prob. K.21ECh. F - Prob. K.22ECh. F - Prob. K.23ECh. F - Prob. K.24ECh. F - Prob. K.25ECh. F - Prob. K.26ECh. F - Prob. L.1ASTCh. F - Prob. L.1BSTCh. F - Prob. L.2ASTCh. F - Prob. L.2BSTCh. F - Prob. L.3ASTCh. F - Prob. L.3BSTCh. F - Prob. L.1ECh. F - Prob. L.2ECh. F - Prob. L.3ECh. F - Prob. L.4ECh. F - Prob. L.5ECh. F - Prob. L.6ECh. F - Prob. L.7ECh. F - Prob. L.8ECh. F - Prob. L.9ECh. F - Prob. L.10ECh. F - Prob. L.11ECh. F - Prob. L.12ECh. F - Prob. L.13ECh. F - Prob. L.14ECh. F - Prob. L.15ECh. F - Prob. L.16ECh. F - Prob. L.17ECh. F - Prob. L.18ECh. F - Prob. L.19ECh. F - Prob. L.20ECh. F - Prob. L.21ECh. F - Prob. L.22ECh. F - Prob. L.23ECh. F - Prob. L.24ECh. F - Prob. L.25ECh. F - Prob. L.29ECh. F - Prob. L.30ECh. F - Prob. L.31ECh. F - Prob. L.32ECh. F - Prob. L.33ECh. F - Prob. L.34ECh. F - Prob. L.35ECh. F - Prob. L.37ECh. F - Prob. L.38ECh. F - Prob. L.39ECh. F - Prob. L.40ECh. F - Prob. L.41ECh. F - Prob. L.42ECh. F - Prob. M.1ASTCh. F - Prob. M.1BSTCh. F - Prob. M.2ASTCh. F - Prob. M.2BSTCh. F - Prob. M.3ASTCh. F - Prob. M.3BSTCh. F - Prob. M.4ASTCh. F - Prob. M.4BSTCh. F - Prob. M.1ECh. F - Prob. M.2ECh. F - Prob. M.3ECh. F - Prob. M.4ECh. F - Prob. M.5ECh. F - Prob. M.6ECh. F - Prob. M.7ECh. F - Prob. M.8ECh. F - Prob. M.9ECh. F - Prob. M.10ECh. F - Prob. M.11ECh. F - Prob. M.12ECh. F - Prob. M.13ECh. F - Prob. M.14ECh. F - Prob. M.15ECh. F - Prob. M.16ECh. F - Prob. M.17ECh. F - Prob. M.18ECh. F - Prob. M.19ECh. F - Prob. M.20ECh. F - Prob. M.21ECh. F - Prob. M.22ECh. F - Prob. M.23ECh. F - Prob. M.25ECh. F - Prob. M.26ECh. F - Prob. M.27ECh. F - Prob. M.28E
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