0:02 this is msj CM in this video I'll be
0:05 looking at avagadro's Law avagadro's law
0:08 was covered in a previous video and we
0:10 saw that the volume occupied by a gas is
0:12 directly proportional to the amounted
0:15 moles of gas at constant pressure and
0:17 temperature this can be shown as V is
0:21 directly proportional to n v/ n equals a
0:26 constant K or V1 / N1 is equal to V2
0:29 over N2 avagadro's law can also be
0:31 expressed as the following at the same
0:34 temperature and pressure equal volumes
0:37 of any gas contain the same number of
0:39 particles so this tells us that for any
0:41 gas equal volumes at the same
0:43 temperature and pressure will contain
0:46 the same number of particles so next we
0:48 look at some examples in this table we
0:51 have the gases oxygen hydrogen nitrogen
0:54 carbon dioxide and methane as we can see
0:56 one mole of each gas has a volume of
1:01 22.7 cubic decim under condition of STP
1:05 which is 22.7 cubic decim if we look at
1:07 the right column we can see that one
1:09 mole of each gas contains the same
1:14 number of particles which is 6.02 * 10 ^
1:17 23 so from this table we can see that
1:19 equal volumes of any gas at the same
1:22 temperature and pressure have the same
1:24 number of particles which is avagadro's
1:27 law so next we look at some example
1:30 problems in the first example we we'll
1:31 look at the reaction between carbon
1:34 monoxide and oxygen to produce carbon
1:36 dioxide according to the balanced
1:40 equation 2 moles of carbon monoxide
1:43 react with one mole of oxygen to produce
1:45 2 moles of carbon dioxide if the
1:47 reaction is carried out at a fixed
1:49 temperature and pressure we can treat
1:53 the mole ratios as volume ratios so we
1:55 can say that two volumes of carbon
1:57 monoxide will react with one volume of
2:01 oxygen to produce two volumes of carbon
2:04 dioxide so 2 cubic decim of carbon
2:06 monoxide will react with 1 cubic decim
2:09 of oxygen to produce 2 cubic decim of
2:13 carbon dioxide and 10 cubic decim of
2:15 carbon monoxide will react with 5 cubic
2:18 decim of oxygen to produce 10 cubic
2:21 decim of carbon dioxide so to summarize
2:23 if a reaction is carried out at a fixed
2:26 temperature and pressure we can treat
2:30 the mole ratios as volume ratios so next
2:32 we look at some examples in our first
2:35 example 40 cubic decim of carbon
2:38 monoxide is reacted with 40 cubic decim
2:40 of oxygen at
2:43 STP determine the volume of carbon
2:46 dioxide produced so the first step is to
2:48 determine the limiting reactant to do
2:50 this we can divide the volume of each
2:52 reactant by its coefficient in the
2:55 balanced equation so for carbon monoxide
2:59 that's 40 / 2 which gives us 20 and for
3:01 oxygen it's 40 IDE by 1 which gives us
3:04 40 the lowest value which is carbon
3:07 monoxide is the limiting reactant and
3:10 oxygen is the excess reactant if we look
3:11 at the coefficients in the balanced
3:13 equation we can see that 2 moles of
3:16 carbon monoxide will produce 2 moles of carbon
3:17 carbon
3:20 dioxide so 40 cubic decim of carbon
3:23 monoxide will produce the same volume
3:26 which is 40 cubic decim of carbon
3:29 dioxide in our next example we have 20
3:32 cub decim of nitrogen reacting with 50
3:36 cubic decim of hydrogen at STP determine
3:38 the volume of ammonia produced and the
3:41 volume of the excess reactant remaining
3:42 so in this reaction which is known as
3:46 the harbor process nitrogen and hydrogen
3:48 react together to form ammonia like in
3:50 the previous example the first step is
3:53 to determine the limiting reactant so
3:55 for nitrogen we divide the volume by the
3:57 coefficient in the balanced equation
4:00 which gives us 20 if we do the same for
4:02 hydrogen we get
4:05 16.7 which tells us that hydrogen is the
4:07 limitin reactant if we look at the molar
4:10 ratio of hydrogen to ammonia we can see
4:13 it's a 3:2 ratio this means that 50
4:16 cubic decim of hydrogen will produce 2/3
4:20 as much ammonia which is 33.3 cubic
4:23 decim next we'll calculate the volume of
4:25 the acccess reactant remaining to do
4:27 this we look at the molar ratio of
4:29 nitrogen to hydrogen which we can see is
4:33 a 1:3 ratio this tells us that 50 cubic
4:35 decim of hydrogen will react with a
4:39 third as much nitrogen which is 16.7
4:42 cubic decim so to calculate the volume
4:45 of nitrogen remaining we subtract 16.7
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