Balancing Equations
Here are a couple of final examples that take a bit of
work to get.
Let’s try another example:
? NH3(g)
( ) + ? O2(g)
( ) = ? NO(g)
NO( ) + ? H2O(g)
O( )
N
1
0
1
0
I usually try to leave H for last, and O for second to last,
so let’s try the N first.
One N in the NH3 on the left,, one N in the NO on the right,
g ,
I can set both coefficients to 1.
1 NH3(g)
( ) + ? O2(g)
( ) = 1 NO(g)
NO( ) + ? H2O(g)
O( )
N
O
1(1)
0
0
2
1(1)
1(1)
0
1(1)
Now let’s work on the O’s
O’s.
O’s look good, so let’s lock down the coefficient of O2 and
H2O at 1.
1 NH3(g)
( ) + 1 O2(g)
( ) = 1 NO(g)
NO( ) + 1 H2O(g)
O( )
N
O
H
1(1)
0
3
0
1(2)
0
1(1)
1(1)
0
0
1(1)
2
So, let’s try to finish with the H’s
Oh Phooey. H doesn’t balance. Not only that, but you can’t
change a single coefficient on either side of the equation to
make
k it balance.
b l
What
Wh t we have
h
to
t do
d now is
i to
t cross multiply,
lti l
that is, use number of H’s in H2O (2) as the coefficient for NH3,
and use the number of H’s in NH3 (3) as the coefficient for H2O.
And, since this messes up one the the coefficients we worked on
before, we have to start the whole process over.
H
N
2 NH3(g)
( ) + ? O2(g)
( ) = ? NO(g)
NO( ) + 3 H2O(g)
O( )
2(3)
0
0
3(2)
2(1)
( )
0
1
0
Try the N’s again.
So we will need a coefficient of 2 in front of the NO.
H
N
O
2 NH3(g)
( ) + ? O2(g)
( ) = 2 NO(g)
NO( ) + 3 H2O(g)
O( )
2(3)
0
0
3(2)
2(1)
( )
0
2(1)
( )
0
0
2
2(1)
3(1)
Try the O’s
O s again.
Wow. This just keeps getting worse. I have a total of 5 O’s on
the right. If I stick a coefficient of 2 in front of the O2 I get 4 on
the right, if I use a coefficient of 3 I get 6 on the left, so neither
works!
It looks like I have to cross multiply again. I will use 5, the
total number of O’s on the left as my coefficient for O2, and I
will use 2, the number of O
O’ss in O2 to multiply the coefficients
of BOTH NO and H2O. And, as before, since I am messing with
coefficients that were set, I have to start over.
? NH3(g)
( ) + 1 O2(g)
( ) = 2 NO(g)
NO( ) + 3 H2O(g)
O( )
x5
[2 NO(g) + 3 H2O(g)] x2
? NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(g)
And, just to check that I didn’t mess up, let’s count O’s again
O
10
4
6
So O looks balanced, and I will start from the point
? NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(g)
on the next page.
O
N
? NH3(g)
( ) + 5 O2(g)
( ) = 4 NO(g)
NO( ) + 6 H2O(g)
O( )
5(2)
4(1)
6(1)
1
0
4(1)
0
Let’ss try the N’s
Let
N s again
again.
So the coefficient on NH3 will have to be a 4.
O
N
H
4 NH3(g)
( ) + 5 O2(g)
( ) = 4 NO(g)
NO( ) + 6 H2O(g)
O( )
5(2)
4(1)
6(1)
4(1)
0
4(1)
0
4(3)
0
0
6(2)
And finally the H’s…….cross your finger’s
Hooray, they all balance!!!!
This was a particularly ugly example, but it just goes to
show you that if you take a consistent approach,
and
d keep
k
after
ft it,
it you can eventually
t ll gett it tto work.
k
Balancing Equations
Let that last problem be a lesson; You don’t always get a
g pass
p
through
g the elements.
final answer on a single
Sometimes you have to work through a couple of times to
get everything to work out.
Let’s try a one more example.
C 3FeS
Cu
F S3(s)
( ) + O2(g)
( ) = Cu(s)
C ( ) + FeO(s)
F O( ) + SO2(g)
( )
Cu
3
0
1
0
0
When I have a big complicated molecule I start with him,
and then work all the elements in that compound first,
so let’s start with Cu
It looks like I will need 3 Cu(s).
1 Cu
C 3FeS
F S3(s)
( ) + O2(g)
( ) = 3 Cu(s)
C ( ) + FeO(s)
F O( ) + SO2(g)
( )
Cu
Fe
1(3)
1(1)
0
0
Now try the Fe.
I can set FeO at 1.
3(1)
0
0
1
0
0
1 Cu
C 3FeS
F S3(s)
( ) + O2(g)
( ) = 3 Cu(s)
C ( ) + 1 FeO(s)
F O( ) + SO2(g)
( )
Cu
Fe
S
1(3)
1(1)
1(3)
0
0
0
Now try the S.
I can set SO2 at 3.
3(1)
0
0
0
1(1)
0
0
0
1
1 Cu
C 3FeS
F S3(s)
( ) + O2(g)
( ) = 3 Cu(s)
C ( ) + 1 FeO(s)
F O( ) + 3 SO2(g)
( )
Cu
Fe
S
O
1(3)
1(1)
1(3)
0
0
0
0
2
3(1)
0
0
0
0
1(1)
0
1
0
0
3(1)
3(2)
And finally the O.
Another nasty. 7 O’s on the right, 2 on the left. Let’s try
another cross multiplication.
multiplication The sum of the O
O’ss on the right
is 7, so we’ll multiply the O2 by 7, and we’ll multiply BOTH
the FeO and the SO2 by 2, the number of O’s in O2. As usual,
since this messes up a coefficient that we had set, we have
to start over as well.
? Cu
C 3FeS
F S3(s)
( ) + O2(g)
( ) = ? Cu(s)
C ( ) + 1 FeO(s)
F O( ) + 3 SO2(g)
( )
x7
(
)x2
? Cu3FeS3(s) +7 O2(g) = ? Cu(s) + 2 FeO(s) + 6 SO2(g)
Double check that the O count is right
right.
O
7(2)=14
14
=
2(1)=2
14
So let’s start from there on the next page.
6(2)=12
? Cu
C 3FeS
F S3(s)
( ) + 7 O2(g)
( ) = ? Cu(s)
C ( ) + 2 FeO(s)
F O( ) + 6 SO2(g)
( )
O
Fe
1
7(2)
0
0
2(1)
2(1)
6(2)
0
Since the FeO coefficient on the right is set, do Fe on the
l f next.
left
So it looks like we need 2 Cu3FeS3 on the left.
2 Cu
C 3FeS
F S3(s)
( ) + 7 O2(g)
( ) = ? Cu(s)
C ( ) + 2 FeO(s)
F O( ) + 6 SO2(g)
( )
O
Fe
S
Cu
2(1)
2(3)
2(3)
7(2)
0
0
0
0
0
1
2(1)
2(1)
0
0
6(2)
0
6(1)
0
Since the SO2 coefficient on the right is set, check that the
S is okk on the
h left
l f next.
S looks good, so let’s get the Cu taken care of.
It llooks
k like
lik we need
d 6 Cu’s
C ’ on the
th right,
i ht and
d that
th t should
h ld do
d it.
it
2 Cu
C 3FeS
F S3(s)
( ) + 7 O2(g)
( ) = 6 Cu(s)
C ( ) + 2 FeO(s)
F O( ) + 6 SO2(g)
( )
O
Fe
S
Cu
2(1)
2(3)
2(3)
7(2)
0
0
0
0
0
6(1)
2(1)
2(1)
0
0
6(2)
0
6(1)
0
And that is the final answer:
2 Cu3FeS3(s) + 7 O2(g) = 6 Cu(s) + 2 FeO(s) + 6 SO2(g)
Balancing Equations
Examples to try:
Zn(s) + HCl(aq) = ZnCl2(aq) + H2(g)
KClO3(s) = KCl(s) + O2(g)
CaSiO3(s) + HF(aq) = CaF2(aq) + SiF4(g) + H2O(l)
Cr(s) + S8(s) = Cr2S3(s)
(Answers on next slide, work the problems and then advance
to the last slide and see if you were right)
Balancing Equations
Zn(s) + HCl(aq) = ZnCl2(aq) + H2(g)
( ) + 2 HCl(aq)
( q) = 1 ZnCl2((aq)
q) + 1 H2(g)
1 Zn(s)
KClO3(s) = KCl(s) + O2(g)
2 KClO3(s) = 2 KCl(s) + 3 O2(g)
CaSiO3(s) + HF(aq) = CaF2(aq) + SiF4(g) + H2O(l)
1 CaSiO
C S O3(s)
( ) + 6 HF(aq)
( )=
1 CaF2(aq) + 1 SiF4(g) + 3 H2O(l)
Cr(s) + S8(s) = Cr2S3(s)
16 Cr(s) + 3 S8(s) = 8 Cr2S3(s)