S1.4.6 Avogadro's law | Mike Sugiyama Jones (MSJ Chem) | YouTubeToText
YouTube Transcript: S1.4.6 Avogadro's law
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this is msj CM in this video I'll be
looking at avagadro's Law avagadro's law
was covered in a previous video and we
saw that the volume occupied by a gas is
directly proportional to the amounted
moles of gas at constant pressure and
temperature this can be shown as V is
directly proportional to n v/ n equals a
constant K or V1 / N1 is equal to V2
over N2 avagadro's law can also be
expressed as the following at the same
temperature and pressure equal volumes
of any gas contain the same number of
particles so this tells us that for any
gas equal volumes at the same
temperature and pressure will contain
the same number of particles so next we
look at some examples in this table we
have the gases oxygen hydrogen nitrogen
carbon dioxide and methane as we can see
one mole of each gas has a volume of
22.7 cubic decim under condition of STP
which is 22.7 cubic decim if we look at
the right column we can see that one
mole of each gas contains the same
number of particles which is 6.02 * 10 ^
23 so from this table we can see that
equal volumes of any gas at the same
temperature and pressure have the same
number of particles which is avagadro's
law so next we look at some example
problems in the first example we we'll
look at the reaction between carbon
monoxide and oxygen to produce carbon
dioxide according to the balanced
equation 2 moles of carbon monoxide
react with one mole of oxygen to produce
2 moles of carbon dioxide if the
reaction is carried out at a fixed
temperature and pressure we can treat
the mole ratios as volume ratios so we
can say that two volumes of carbon
monoxide will react with one volume of
oxygen to produce two volumes of carbon
dioxide so 2 cubic decim of carbon
monoxide will react with 1 cubic decim
of oxygen to produce 2 cubic decim of
carbon dioxide and 10 cubic decim of
carbon monoxide will react with 5 cubic
decim of oxygen to produce 10 cubic
decim of carbon dioxide so to summarize
if a reaction is carried out at a fixed
temperature and pressure we can treat
the mole ratios as volume ratios so next
we look at some examples in our first
example 40 cubic decim of carbon
monoxide is reacted with 40 cubic decim
of oxygen at
STP determine the volume of carbon
dioxide produced so the first step is to
determine the limiting reactant to do
this we can divide the volume of each
reactant by its coefficient in the
balanced equation so for carbon monoxide
that's 40 / 2 which gives us 20 and for
oxygen it's 40 IDE by 1 which gives us
40 the lowest value which is carbon
monoxide is the limiting reactant and
oxygen is the excess reactant if we look
at the coefficients in the balanced
equation we can see that 2 moles of
carbon monoxide will produce 2 moles of carbon
carbon
dioxide so 40 cubic decim of carbon
monoxide will produce the same volume
which is 40 cubic decim of carbon
dioxide in our next example we have 20
cub decim of nitrogen reacting with 50
cubic decim of hydrogen at STP determine
the volume of ammonia produced and the
volume of the excess reactant remaining
so in this reaction which is known as
the harbor process nitrogen and hydrogen
react together to form ammonia like in
the previous example the first step is
to determine the limiting reactant so
for nitrogen we divide the volume by the
coefficient in the balanced equation
which gives us 20 if we do the same for
hydrogen we get
16.7 which tells us that hydrogen is the
limitin reactant if we look at the molar
ratio of hydrogen to ammonia we can see
it's a 3:2 ratio this means that 50
cubic decim of hydrogen will produce 2/3
as much ammonia which is 33.3 cubic
decim next we'll calculate the volume of
the acccess reactant remaining to do
this we look at the molar ratio of
nitrogen to hydrogen which we can see is
a 1:3 ratio this tells us that 50 cubic
decim of hydrogen will react with a
third as much nitrogen which is 16.7
cubic decim so to calculate the volume
of nitrogen remaining we subtract 16.7
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