Practice Problem: Balancing Equations

Practice Problem: Balancing Equations


okay so let’s go
through a problem regarding balancing equations, so here are four equations that we’re going to need to balance if you’ll
notice they are not currently balanced and so this is pretty straightforward
there’s nothing much more to say if you need any help with this concept and some
strategies that you can use in order to do this check out my tutorial on
balancing equations otherwise when you’re ready go ahead and give these a
try. so let’s start with this first one we have CaCl2 plus Na3PO4
yields Ca3(PO4)2 plus NaCl so if you have trouble with these and you want to
be very very thorough in your approach the best ways to make a table like this
where we’ve listed every element that is involved in the reaction and then we can
tally the number of atoms of each element on the left side and on the
right side. so let’s go ahead and do that now, we have got one calcium atom on the
left and three calcium atoms on the right. for chlorine we have two on the
left because we have Cl2 and on the right we have just one in NaCl. for
sodium we have 3 on the left and we have one on the right. for phosphorus we have
one on the left and then we have two on the right, don’t forget that with PO4 in
parenthesis then we have the two outside of the parenthesis that means there are
two phosphate polyatomic ions so it’s PO4 twice that means we have PO4 and
another PO4 so that’s two phosphorus atoms. and then four oxygens, same thing
on the left we have four on the right we actually have eight because we have O4 but PO4 is twice so we have O4 and
another O4 that’s 8, that’s 8 oxygen atoms so that’s the number
of atoms of each element that we have on each side and now all of those need to
be balanced. so where shall we start let’s just go ahead and start with
calcium. so all we need is a coefficient to the left of some substance to balance
the number of atoms of that particular element to get the same number on each
side. so for calcium we have one on the left and three on the right, so the
easiest thing to do will be to put a three right here. so now we have CaCl2
but that coefficient of three means we have three of those so we have CaCl2,
CaCl2 and another of CaCl2. so that means we have one calcium for
each of those so we have three calcium atoms total. so let’s change our tally
for calcium on the left side to three so calcium is now balanced, we have three on
the left and three on the right so that’s done but what we must realize is
that we also have made it so that there are six chlorine atoms on the left,
because we have Cl2 but we have three of those so two chlorine atoms per
formula unit times three of those means there are six chlorine atoms on the left
now. so what shall we do next? so we’ll come back to chlorine but what we want
to understand is that there’s a simpler way to do this than to do phosphorus and
oxygen separately because phosphorus and oxygen are involved in the phosphate ion.
it’s a polyatomic ion and those are not going to get split up they remain
together so we can actually balance the entire phosphate ion at the same time so
we take a look at the left we have a PO4 and then on the right we have (PO4)2 so
we can treat that as though it’s an individual thing we can just try to get
the same number of phosphate ions on both sides so as we said we have two
phosphate ions on the right so if we put a two here that means we are gonna have
two phosphate ions on the left and now look at what we’ve done we have made it
so that there are two phosphorus atoms on the left and eight oxygen atoms on
the left because we have Na3PO4 we have two of those so all
of those are multiplied by two and now both phosphorus and oxygen are balanced.
however we have also altered the number of sodium atoms on the left side because
we have Na3 and we have multiplied that by 2 so we need to change the number of
sodium atoms on the left side to 6 so what should we do next well as it turns
out we can go ahead and take care of both chlorine and sodium at the same
time because on the right we have NaCl it looks like we need to put a 6 here
because we have six sodiums on the left and six chlorines on the left and only
one of each on the right so if we put a six there we are going to change the
number of sodium and chlorine atoms on the right both to six and so that is
going to balance both of those elements and that will actually completely
balance the equation so those are the coefficients in the balanced equation
for this reaction. okay let’s look at the next one we have a combustion reaction
we have C2H6O plus O2 yields CO2 plus H2O, so let’s make our table. it’s a
little bit smaller all we have is carbon hydrogen and oxygen so let’s make our
initial tally for carbon. we have two carbon atoms on the left and one carbon
atom on the right. for hydrogen we have six atoms on the left and we have two on
the right. and for oxygen we have three on the left because we have one oxygen
atom in C2H6O and we have an O2 molecule and then on the right we have also three
because we have O2 in the CO2 and we have one oxygen atom in water so that is
our initial tally. so where should we start, well here is one handy tip anytime
you have some element that is by itself in a particular species you want to do
that one last because you have the most freedom with that one you can get
everything else balanced and then you can just put whatever
coefficient you want for that element to make it work without affecting the
numbers for any other element. so we are gonna do oxygen last also because
currently it is balanced so we’re gonna have to mess with the other two elements
first but in general if you see O2 or N2 or H2 you can feel free to do that one last.
so why don’t we start with carbon, we have two on the left and one on the
right so what we need to do is put a 2 next to CO2 so that we have two carbon
atoms on the right so now carbon is balanced. however we have also changed
the number of oxygen atoms on the right we have four from CO2 because we have
two CO2 so there’s a CO2 and another CO2 each of
those has two oxygen atoms so that’s four oxygen atoms from CO2 plus the one
from water and that gives us five oxygen atoms on the right. now let’s balance
hydrogen we have six on the left and we have two on the right so we need to put
a three next to water because that is going to give us three water molecules
each of which has two hydrogen atoms that gives us a total of six hydrogen
atoms on the right and that will balance hydrogen, however we have also again
changed the number of oxygen atoms because we now have four from CO2 plus
three from water because we have three waters for a total of seven oxygen atoms
on the right so lastly we can balance oxygen now we have seven on the right
and on the left we have one from C2H6O and we’re probably not going to want to
mess with that because that will alter the number of carbons and hydrogens and
so what we want to do is manipulate O2. so if we need to get seven on the left
and one of them is in C2H6O that leaves a total of six that we need from O2.
now luckily that’s going to be very easy to do because all we have to do is
put a 3 here and now we have 3O2 molecules so an O2 + O2 and another O2
and that adds up to six oxygen atoms so the six oxygens from oO plus the other
one gives us 7 on the left and now oxygen is balanced and the equation is
balanced in general. let’s move on to our third example we have CO + H2 yields
C8H18 + H2O. so let’s make our table again once again we just have C, H and O
so let’s make our tally. carbon we’ve got one carbon atom on the left and we have
eight carbon atoms on the right from C8. for hydrogen we have just 2 on the left
and on the right we have 20 because we have H18 and we have each 2 from water
so 18 plus 2 is 20. and then for oxygen we have just the one from CO and the one
from H2O. so once again remember that last trick that we learned if we have
one element that’s by itself we want to do that one last and in this case that’s
hydrogen so we want to save hydrogen for last.
now oxygen happens to be balanced at the moment so let’s start with carbon. for
carbon we’ve got eight carbon atoms on the right and we have just one on the
left so we’re gonna need to put an eight here so that we have eight CO molecules
and therefore eight carbon atoms on the left. and now carbon is balanced, however
we have also altered the number of oxygen atoms on the left side. we now
have eight of those so now let’s go ahead and balance oxygen. since we have
disrupted that balance we have eight on the left and we have one on the right so
let’s put an eight here and now we have eight oxygen atoms on both sides and
oxygen is balanced now let’s do hydrogen on the right side we have 18 from C8H18
and we have 16 from water because we have two hydrogen atoms in each water
molecule times eight water molecules and 18 plus 16 is 34 so we have 34 on the
right side but this is the easy part because on the left we have H2 by itself
so all we have to do is put a coefficient next to H2 that will make
this balanced and it won’t affect any other element. so 34 divided by 2 is 17,
that means we need 17 H2 molecules. that will give us 34 hydrogen atoms on the
left and hydrogen is now balanced and the equation is balanced okay one more
NH4NO3 yields N2 + O2 + H2O. once again we make our table this time it has
N H and O, for nitrogen we have 2 on the left and two on the right, for hydrogen
we have 4 on the left and two on the right, and for oxygen we have 3 on the
left and 3 on the right. so it seems that everything is balanced except hydrogen
so let’s go ahead and work with that first. we have 4 on the left and two on
the right so we need to put a 2 next to water because two water molecules will
yield 4 hydrogen atoms so now we have 4 on the right and hydrogen is balanced
however we have changed the number of oxygen atoms on the right, we have 2 from
O2 and now we have 2 from water, one per water molecule and 2 water molecules so
now we have 4 oxygen atoms on the right so let’s go ahead and balance oxygen now
here’s where it gets a little bit trickier because we have 4 on the right
and we have 3 on the left so there’s nothing we can immediately do to balance
it but we know that we have more oxygen on the right than we do on the left and
we only have one thing on the left that we can work with so let’s just go ahead
and put a 2 here because we need more on the left but this is the best we can
do so let’s just put a two here and that will give us six oxygen atoms on the
left so now we have something to work with on the right, we can we can do some
things to balance oxygen a little bit later but we needed to do that to get
more oxygen on the left. however we have now changed some other things we now
have eight hydrogen atoms on the left so unfortunately hydrogen is no longer
balanced. we also have four nitrogen atoms on the left now because each of
those has two nitrogen atoms and now we have two of them so we have four
nitrogen atoms on the left so now let’s start balancing again. let’s go to oxygen
one more time we have six on the left and we have four on the right so what is
one way that we could get two more oxygen atoms? well we have oxygen in
water so to get two more oxygens we could just have two more waters so let’s
change this to four waters and that gives us six oxygen atoms on the right
so oxygen is balanced. it also gives us eight hydrogen atoms on the right which
means that hydrogen is also balanced so that was a good step. lastly we just need
to do nitrogen, luckily nitrogen is by itself on the right side so to get four
nitrogen atoms on the right side we just put a two here so that we have 2N2 that
gives us four nitrogens on the right so now nitrogen is balanced and the
equation is balanced.

8 thoughts on “Practice Problem: Balancing Equations

  • January 22, 2019 at 5:21 pm
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    first? nice video! <3 ur channel ;D

    Reply
  • January 22, 2019 at 5:23 pm
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    Please make vedios on various kinds of pericyclic reactions

    Reply
  • January 22, 2019 at 5:26 pm
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    I have done all correctly in 4 min

    Reply
  • January 22, 2019 at 11:07 pm
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    Thank you sir

    Reply
  • January 22, 2019 at 11:57 pm
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    So helpful

    Reply
  • January 23, 2019 at 5:29 am
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    Bless

    Reply
  • January 23, 2019 at 12:20 pm
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    Professus Davus Explanus

    Reply
  • July 2, 2019 at 2:08 pm
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    Can I balance the last one by doing 1, 1, 1/2, 2 (instead of 2, 2, 1, 4)?

    Reply

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