hello and welcome to the chemistry
solution this tutorial is on balancing
combustion reactions for the purposes of
this tutorial will focus on hydrocarbon
combustion and that means will combust
hydrocarbons which are compounds that
contain only carbon and hydrogen and
we'll assume that these react with o2 in
the air to form co2 which is carbon
dioxide and h2o which is water we'll
also assume complete combustion which
means that the only products formed will
be co2 and h2o let's start by balancing
the following equation the first thing
we need to do is to take an inventory of
how many of each type of atom we have on
the reactant side and the product side
on the reactant side we have five carbon
atoms 10 hydrogen atoms and two oxygen
atoms well on the product side we have
one carbon atom two hydrogen atoms and
three oxygen atoms I think that it's
always helpful to start balancing
combustion reactions by balancing the
number of carbon atoms first and we need
to remember that when you balance
equations you can only add coefficients
in front of your molecules you can never
change the subscript because if you
change the subscript you change the
identity of the molecule now because we
have five carbon atoms on the reactant
side we need to put a five in front of
co2 on the product side to give us five
carbon atoms on the product side but by
putting a five in front of co2 we've not
only changed the number of carbon atoms
on the product side to five but we've
also changed the number of oxygen atoms
on the product side so we have ten
oxygen atoms from co2 and one oxygen
atom from h2o giving us a total of
eleven oxygen atoms now let's balance
the number of hydrogen atoms and the
reason I'm choosing to balance hydrogen
first before oxygen is because hydrogen
occurs and only one compound on the
reactant side and one compound on the
product side so
because we have ten hydrogen atoms on
the reactant side and two on the product
side we'll need to put a five
coefficient in front of h2o to give us
ten hydrogen atoms on the product side
but in putting a five in front of h2o
we've not only changed the number of
hydrogen atoms on the product side but
we've changed the number of oxygen atoms
as well so we now have ten oxygen atoms
from co2 which we found by taking five
times the subscript on oxygen to to give
us 10 but we have an additional five
oxygen atoms from h2o giving us a total
number of oxygen atoms of 15 now we need
to balance the number of oxygen atoms
now you'll see that we have o to on the
reactant side and we have 15 oxygen
atoms on the product side that we need
to balance now putting any coefficient
in front of o2 is going to give us an
even number of oxygen atoms and so the
trick to balancing these combustion
reactions is to start by putting a
fractional coefficient in front of otwo
so in this case we'll choose 15 halves
because 15 halves times the coefficient
on Oh 2 which is 2 gives us 15 oxygen
atoms on the reactant side which now
balances with the 15 oxygen atoms on the
product side but we can't have a
balanced equation with fractions as
coefficients so we need to multiply all
of the coefficients in our entire
reaction by 2 so that we have all whole
number coefficients and when we do that
we'll see that we have to c 5h 10 plus
15 o 2 gives us 10 co2 plus 10 h2o and
if you count up the number of carbon
hydrogen and oxygen atoms on your
reactant side they will equal the number
of carbon hydrogen and oxygen atoms on
your product side let's try another
problem in this reaction
we're combusting hexane which has the
formula c6h4 teen in air and the
molecule in air that's going to react in
this combustion reaction
it'sjust Oh - now hexane and o2 will
combust to give us co2 and h2o so first
let's take inventory of how many of each
different type of atom we have on our
reactant side and our product side and
then let's start by balancing the carbon
atoms first so to balance out the six
carbon atoms on our reactant side we
need to put the coefficient six in front
of co2 that now gives us six carbon
atoms on our product side but it also
increases the number of oxygen atoms on
our product side so we have 12 oxygen
atoms from the six co2 molecules plus
one oxygen atom from h2o let's balance
the number of hydrogen atoms next so we
have 14 hydrogen atoms on our reactant
side and so in order to have 14 hydrogen
atoms on our product side we need to put
the coefficient seven in front of h2o
that gives us 7 times 2 14 hydrogen
atoms but it also increases the number
of oxygen atoms we have on our product
side so now we have 12 oxygen atoms from
our six co2 molecules plus an additional
7 oxygen atoms from our 7 molecules of
h2o for a total of 19 oxygen atoms now
just like in our last example we have o
to on our reactant side so any whole
number coefficient we put in front of
otwo will give us an even number of
oxygen atoms and we have 19 oxygen atoms
that we're trying to balance on the
product side so let's start by putting a
fraction coefficient in front of o2 19
halves 19 1/2 times 2 will give us 19
oxygen atoms so now the number of oxygen
is balanced but we can't have a
fractional coefficient in our balanced
equation so we need to multiply the
entire equation by 2 so that gives us 2
c6h14
plus 1902 gives us 12 co2 plus 14 h2o
and when we tabulate the number of
carbon hydrogen and oxygen atoms you'll
see that we have 12 carbon atoms 28
hydrogen atoms and 38 oxygen atoms on
both the reactant side and the product
side and along with the fact that all of
the coefficients in our balanced
equation are in their lowest ratio means
that this equation is correctly balanced
thanks for watching the chemistry
solution we hope you enjoyed this
tutorial
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