Balancing Equations. Balancing a chemical equation refers to the skill (or art) of

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Balancing Equations

Balancing a chemical equation refers to the skill (or art) of making sure that all atoms that are written on the left of a making sure that all atoms that are written on the left of a chemical reaction as reactants are also present on the right side of the equation as products.

Since many of the problems you will see in this class require you to start with a balanced chemical equation, this is a skill you must learn now and will use over and over of the course of the year.

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Balancing Equations

Balancing a chemical equation isn’t that hard to do, it is a simple trial-and-error process that is repeated for each

simple trial and error process that is repeated for each element in the equation until all elements are balanced.

After you get the hang of it it can even be kind of fun as you

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Balancing Equations

Example 1. Balance the following chemical equation: FeCl3(aq) + AgNO3(aq) = AgCl(s) + Fe(NO3)3(aq) FeCl3(aq) + AgNO3(aq) = AgCl(s) + Fe(NO3)3(aq) The (aq) and the (s) refer to the physical form of the

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chemical and we will talk about that more in chapter 4, for now the physical descriptors can be ignored.

To balance the equation the only change you can make is to change the coefficient in front of a molecule. You cannot

change any of the subscripts in a molecule because that

change any of the subscripts in a molecule because that

changes the chemical to a different substance.

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? FeCl33( q)(aq) + ? AgNOg 33(aq) = ? AgCl(s) + ? Fe(NO( q) g ( ) ( 33))33(aq)( q)

Fe 1 0 0 1

So pick an element and make sure you have the same So pick an element and make sure you have the same

number of atoms of that element on both sides of the equation. For instance, let’s start with the iron (Fe). How many Fe’s

are there on each side of the equation?

One Fe both sides of the equations so it is already balanced! That means the coefficients for FeCl3 and Fe(NO3)3 can be set to 1.

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1 FeCl33(aq) + ? AgNO( q) g 33( q)(aq) = ? AgCl(s) + g ( ) 1 Fe(NO( 33))33(aq)( q)

Fe 1 0 0 1

So pick an element and make sure you have the same So pick an element and make sure you have the same

number of atoms of that element on both sides of the equation. For instance, let’s start with the iron (Fe). How many Fe’s

are there on each side of the equation?

One Fe both sides of the equations so it is already balanced! That means the coefficients for FeCl3 and Fe(NO3)3 can be set to 1.

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1 FeCl33(aq) + ? AgNO( q) g 33( q)(aq) = ? AgCl(s) + g ( ) 1 Fe(NO( 33))33(aq)( q)

Fe 1 0 0 1

Cl 3 0 1 0

Now let’s try the Cl (Notice I will continue with the table

Cl 3 0 1 0

Now let’s try the Cl. (Notice I will continue with the table that I started on the last page)

3 Cl’s on the left, 1 Cl on the right.

So we need to multiply the AgCl on the right hand side by a p y g g y coefficient of 3 to get 3 Cl’s on this side.

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1 FeCl33(aq) + ? AgNO( q) g 33( q)(aq) = ? AgCl(s) + g ( ) 1 Fe(NO( 33))33(aq)( q)

Fe 1 0 0 1

Cl 3 0 3(1) 0

Now let’s try the Cl (Notice I will continue with the table

Cl 3 0 3(1) 0

Now let’s try the Cl. (Notice I will continue with the table that I started on the last page)

3 Cl’s on the left, 1 Cl on the right.

So we need to multiply the AgCl on the right hand side by a p y g g y coefficient of 3 to get 3 Cl’s on this side.

A d thi t th ffi i t f A Cl t 3 And this sets the coefficient of AgCl to 3

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1 FeCl33(aq) + ? AgNO( q) g 33( q)(aq) = 3 AgCl(s) + g ( ) 1 Fe(NO( 33))33(aq)( q)

Fe 1 0 0 1

Cl 3 0 3(1) 0

Now let’s try the Cl (Notice I will continue with the table

Cl 3 0 3(1) 0

Now let’s try the Cl. (Notice I will continue with the table that I started on the last page)

3 Cl’s on the left, 1 Cl on the right.

So we need to multiply the AgCl on the right hand side by a p y g g y coefficient of 3 to get 3 Cl’s on this side.

A d thi t th ffi i t f A Cl t 3 And this sets the coefficient of AgCl to 3

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1 FeCl33(aq) + ? AgNO( q) g 33( q)(aq) = 3 AgCl(s) + g ( ) 1 Fe(NO( 33))33(aq)( q)

Fe 1 0 0 1

Cl 3 0 3(1) 0 Cl 3 0 3(1) 0

Ag 0 1 3(1) 0

Now let’s try the Ag’s

One on the left, three on the right, we will need to multiply The AgNO by three to make it all balance

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1 FeCl33(aq) + ( q) 3 AgNOg 33(aq) =( q) 3 AgCl(s) + g ( ) 1 Fe(NO( 33))33(aq)( q) Fe 1 0 0 1 Cl 3 0 3(1) 0 Cl 3 0 3(1) 0 Ag 0 3(1) 3(1) 0 N 0 3 0 3

Now let’s try the N’s… O 0 9 0 9

The N’s look good, I didn’t have to mess with any coefficients, so let’s finish with the O’s

so let s finish with the O s

The O’s are balanced as well, so we must be done. The final balanced equation is:

balanced equation is:

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Balancing equations

As you balance a reactant and a product that both contain a given element in an equation you try not to have to change given element in an equation, you try not to have to change the coefficient on either of those two compounds again. If you do, then you have to go back and start over. Notice how

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Balancing equations

Step 1

FeCl3(aq) + AgNO3(aq) = AgCl(s) + Fe(NO3)3(aq) FeCl3(aq) + AgNO3(aq) = AgCl(s) + Fe(NO3)3(aq)

Fe 1 0 0 1

N t t t ith ith F Cl F (NO ) f

Now you try not to mess with either FeCl3 or Fe(NO3)3 from this point onward. I will underline them remind myself to try not to touch these coefficients again.

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Balancing equations

Step 2

FeCl3(aq) + AgNO3(aq) = 3AgCl(s) + Fe(NO3)3(aq) FeCl3(aq) + AgNO3(aq) = 3AgCl(s) + Fe(NO3)3(aq)

Fe 1 0 0 1

Cl 3 0 3(1) 0

Now with the Cl, the AgCl coefficient is fixed, so I will underline that one as well.

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Balancing equations

Step 3

FeCl3(aq) + 3AgNO3(aq) = 3AgCl(s) + Fe(NO3)3(aq) FeCl3(aq) + 3AgNO3(aq) = 3AgCl(s) + Fe(NO3)3(aq)

Fe 1 0 0 1

Cl 3 0 3(1) 0

A 0 3(1) 3 0

Ag 0 3(1) 3 0

Now with the Ag, the AgNO33 coefficient is fixed, so I will underline that one as well.

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Balancing equations

Steps 4 and 5

FeCl33( q)(aq) + 3AgNOg 33(aq) = 3AgCl(s) + Fe(NO( q) g ( ) ( 33))33(aq)( q)

Fe 1 0 0 1 Cl 3 0 3(1) 0 Ag 0 (3)1 0 3 Ag 0 (3)1 0 3 N 0 3 0 3 O 0 9 0 9

And fixing the Ag fixes the coefficient of the last compound, so we cross our fingers that the N and O will come into g

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Balancing Equations

Now try this one:

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? Au2S3(s) + ? H2(g) = ? Au(s) + ? H2S(g) Au 2 0 1 0

Au first.

So you have 1 Au2S3 on the left for a total of 2 Au’s, that means you need 2 Au(s) on the right.

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1 Au2S3(s) + ? H2(g) = 2 Au(s) + ? H2S(g) Au 1(2) 0 2(1) 0

S 3 0 0 1

S 3 0 0 1

S next.

So if you have 1 Au2S3 on the left, you’ll need 3 H2S on the right.

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1 Au2S3(s) + ? H2(g) = 2 Au(s) + 3 H2S(g) Au 1(2) 0 2(1) 0

S 1(3) 0 0 3(1)

H 0 2 0 3(2) H next.

So if you have 3 H2S on the left you’ll need 3 H2’s on the right.

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1 Au2S3(s) + 3 H2(g) = 2 Au(s) + 3 H2S(g) Au 1(2) 0 2(1) 0

S 1(3) 0 0 3(1)

H 0 3(2) 0 3(2) And it all worked, your final answer is:

1 A S ( ) 3 H ( ) 2 A ( ) 3 H S( ) 1 Au2S3(s) + 3 H2(g) = 2 Au(s) + 3 H2S(g)

These were a couple of nice, easy examples.

Figure

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