Stoichiometry: Limiting reagent | Chemical reactions and stoichiometry | Chemistry | Khan Academy
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Stoichiometry: Limiting reagent | Chemical reactions and stoichiometry | Chemistry | Khan Academy


Let’s say we have some
ammonia gas. That’s NH3 and it’s a gas,
that’s why the g is in parentheses. And we combine that with some
oxygen, molecular oxygen, it’s also a gas. And that reaction produces
some nitrogen monoxide. NO, monoxide, there’s only
one oxygen there. That’s also a gas, it’s also
called nitric oxide, not to be confused with nitrous oxide. I’ll write nitrous oxide, it’s
N2O, this is laughing gas. Anyway, I don’t want to divert
you too much, we have to focus on the problem. So this is nitric oxide
or nitrogen monoxide. It’s a pollutant, I think
it comes out of some cigarette smokes. I think it’s used in
the body, as well. You could we could look
it up on the internet. There’s your oxide. Ammonia is an important
fertilizer. You know all about oxygen. And this also yields
some water. So, plus some H2O. And we’re told that we’re given
34 grams of ammonia. And we’re given 32
grams of oxygen. This is going to be the
oxygen molecule, O2. So the question is, how many
grams of nitrogen monoxide, or the nitric oxide, are going
to be produced? So how much of just the NO is
going to be produced in grams? So this is a stoichiometry
problem. And so the important thing first
is to just make sure we have a balanced equation before
we even start anything. And lo and behold, we don’t
have a balanced equation. Let’s confirm that it’s
not balanced. Let’s see, we have one nitrogen
here, we have one nitrogen there. That looks balanced so far. And remember, the pattern is
start with the complicated stuff, leave the single atom
molecules for last. Because those you can fix at the
end without missing anything else up. Hydrogen, we have three
hydrogens on the left-hand side. On the right-hand side we
have two hydrogens. So let’s see, how can we have
three hydrogens on the right-hand side? If we multiply this times one
and a half, 1.5, now we have three hydrogens on this side. 1.5 times 2, we have
three hydrogens on the right-hand side. Things are looking good. We have two oxygen on
the left-hand side. How many oxygens do we have
on the right-hand side? We have one oxygen here, and
we have one oxygen in this molecule, but we have one and a
half of the whole molecule. So we have one and a half
oxygens and then we have one more oxygen. So we have two and a half
oxygens on the right-hand side, we only have two
in this molecule. So what do we have to do? How can we get two and
a half oxygens here? Well if we multiply it
by 5/4, or 1.25. 5/4 times 2 is 5/2,
which is 2.5. So now we have 2.5 oxygens. 5/4 times 2 is 2.5. 1 plus 1.5 is also 2.5. Well, we’re not balanced yet. We can’t leave this equation
with just these weird decimal numbers over there. So let’s rewrite it. If we wanted to get rid of all
of these, we can multiply the entire equation by 2. No, not 2. We have to multiply the entire
equation by 4 to get rid of this 4 in the denominator. So we multiply the entire
equation by 4, we have 4 molecules of ammonia. We can even think of it
in terms of moles. Right now I’m thinking of
individual molecules. But we could say we have
4 moles of ammonia. 4 times 6 times 10 to the
23 molecules of ammonia. Either way it all works out. Hopefully, you’re starting to
see the value of moles. Plus 5 molecules, I’ll just
think of it in terms of individual molecules for
now, plus 5 molecules of molecular oxygen. We’re just multiplying
everything by 4, that’s what we’re doing. Yields 4 moles of nitrogen
monoxide. So 4NO plus, we multiplied
both sides by 4, so plus 6 waters. 6 H2O. So there you go, we
got some good practice balancing equations. So let’s go back to the
original problem. We’re given 34 grams
of ammonia. So what we need to figure out
is how many moles of ammonia were we given? So what’s the atomic
mass of ammonia? Not the atomic mass, what’s the
molecular mass of ammonia? We’re dealing with nitrogen
14, so it has a mass number of 14. Hydrogen has a mass
number of 1. So each nitrogen has
a mass of 14. And then the hydrogens each
have a mass of 1. Remember, hydrogen is
kind of strange. It doesn’t have neutrons,
or at least in its most traditional form. So it has just a mass
number of 1. It’s just a proton and an
electron if it’s neutral. So this is 3 times 1. We have three hydrogen atoms. So
the mass, the atomic mass, of one molecule of ammonia
is 14 plus 3 is 17 atomic mass units. Or, another way to write that,
if one molecule of ammonia is 17 atomic mass units,
then 1 mole of ammonia is how many grams? It’s going to be 17 grams. So how many moles of ammonia
are we given? We’re given 34 grams of ammonia,
1 mole is 17 grams. So we’re given 2 moles. 34 is 2 times 17. So this is 2 moles. We’re given 2 moles
of ammonia. Let’s see how much oxygen
we’re given. Or how much of the molecular
oxygen we’ve been given in this case. So let’s see, the mass of just
oxygen by itself is 16. The mass of just the atomic
oxygen, you have to be a little bit careful here because
sometimes people say we have 32 grams of oxygen when
they’re really talking about the molecular oxygen. Well, I guess it doesn’t
matter either way. But, sometimes when they talk
about oxygen you have to make sure whether it is molecular
or atomic oxygen. But the atomic mass number
of oxygen is 16. I can confirm that
by looking at the periodic table down here. So what’s the molecular mass of
the diatomic molecule O2? Well it has 2 oxygen, so it’s
going to be 2 times 16 equals 32 atomic mass units. One molecule of O2 is 32
atomic mass units. Or 1 mole of O2 is
how many grams? Well if one molecule is 32
atomic mass units, then 6 times 10 to the 23 of that
molecule are going to be that many grams. 32 grams. So how
many moles of oxygen have we been given? We’ve been given exactly 32
grams of oxygen, which is exactly 1 mole. So we’ve been given 34 grams of
ammonia, which is 2 moles. Let me write that in a
nice vibrant color. And we’ve been given 1 mole
of the oxygen molecule. Now, when we look at this
reaction, for every 4 moles of ammonia, we need 5
moles of oxygen. Or for every 5 moles of oxygen
we need 4 moles of ammonia. So something doesn’t gel here. Normally, we need more
moles of oxygen than we have ammonia. In the example that we’re
working through, we’ve been given less moles of oxygen
than ammonia. We’ve been given less
oxygen than we need for all of this ammonia. In an ideal world, if we had 2
moles of ammonia, we would need 2.5 moles of oxygen. The ratio of ammonia
to oxygen, let me write that down. In a different color, this
is getting boring. The ratio of ammonia, NH3 to
oxygen in our balanced equation, let me put that in a
square so you know this is the most important part of
what I’m writing. The ratio in this
reaction is 4:5. So if I’m given 2 moles
of ammonia, this is equal to what? If I’m given 2 moles of ammonia,
how many moles of oxygen do I need? I need 2.5 moles of oxygen. Whatever this is,
is that right? Yeah, 4/5. 1.25. If you doubled both of these
numbers, you get 4/5. So I need 2.5 moles. But I don’t have 2.5
moles of oxygen. I only have 1 mole of oxygen. So oxygen is going to
be the limiting reagent in this reaction. I don’t have enough oxygen. I have plenty of ammonia, but I
don’t have enough oxygen to react with it. So this is the limiting
reagent. And I said before, the word
reagent and reactant are used interchangeably. But when people talk about the
limiting ones, they tend to call it the reagent. So oxygen is the limiting
reagent. So we have extra ammonia. So given that we have 1 mole of
oxygen, how many moles of ammonia can I react with that? So this reaction is going to
look something like this. I only have 1 mole of oxygen. So instead of 5O2, I
have to write 1O2. Let me make sure that’s
not of 10. Let me do it in a
different color. I only have 1O2 instead
of 5O2. So how many ammonia are going
to react with that? Well the ratio is 4:5. So I’m going to have
0.8 ammonia. 4 is to 5 as 0.8 is to 1. And so essentially, if I take
this whole equation up here and divide it by 5,
I’ll get what’s actually going to happen. So this divided by
5 is 0.8, nitric oxide or nitrogen monoxide. Plus 6/5 moles of H2O. And so the original question in
the beginning is, how many grams of nitric oxide
are we going to produce, or nitrogen monoxide? So we have 1 mole of oxygen,
0.8 moles of ammonia, and we’re going to produce 0.8 moles
of nitrogen monoxide. Because we only have
one oxygen. So 0.8 moles of nitric oxide
or nitrogen monoxide. So what’s the atomic
mass, the molecular mass, of nitrogen monoxide? Nitrogen has 14 atomic
mass units. Oxygen is 16. We’ve done that before. But you can confirm, your oxygen
is 16, nitrogen is 14. So one molecule of nitrogen
monoxide is equal to 30 atomic mass units. 1 mole or 6.02 times 10 to the
23 molecules of nitrogen monoxide, therefore, will be 30
grams. And how many moles are we producing in
this reaction? Because oxygen was the limiting
reagent, we only had 1 mole of oxygen here. Because of that, we can
only produce 0.8 moles of nitrogen monoxide. So 0.8 moles of nitrogen
monoxide, 1 mole is 30 grams. So 0.8 moles of NO is going to
be equal to 0.8 times 30, which is equal to 24 grams. So
we’re going to be able to produce 24 grams of
nitrogen monoxide. And so you might ask a question,
we’re only using 0.8 moles of ammonia. We were given, in the original
problem, 2 moles of ammonia, so what happens with all
the leftover ammonia? Assuming we mix it really good,
we’re going to literally end up with 1.2 moles of
ammonia just doing nothing at the end. So we’re going to have 24
grams of nitric oxide. And then we used 0.8
moles of ammonia. And we’re going to have left
1.2 moles of ammonia. And if you want, you can figure
out how many of the original grams of
ammonia that is. You just figure out how many
grams 1 mole of ammonia is, it’s 17 grams. And then just
multiply that times 1.2. Anyway, I hope you found
that interesting. If you found this a little bit
confusing, just watch the video again and pause it and
try to solve it yourself. But I think it should start
making a little bit of intuitive sense. I think the hard part is just
the conversions between moles and grams and getting
that part right. And then making sure that you
understand the ratios. You understand this is a 4:5
ratio, there’s always going to be less ammonia than oxygen. And if I only have 1 mole of
oxygen, I’m going to have to have less than 1 mole
of ammonia, even though I was given 2. So oxygen, in this example, was
the rate limiting factor. Let’s say I had 10 moles of
oxygen and 2 moles of ammonia. In that case, ammonia
would have been the rate limiting factor. Because then I would have had
more than enough oxygen. Because for 2 moles of
ammonia, I only need 2.5 moles of oxygen. So 10 moles would have
been overkill. Anyway, see you in
the next video.

About James Carlton

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100 thoughts on “Stoichiometry: Limiting reagent | Chemical reactions and stoichiometry | Chemistry | Khan Academy

  1. if you're teaching someone the concept of a limiting reagents and Stoichiometry for the first time there's really no point in overcomplicating it and throwing in fractions. just saying

  2. I think that to understand everything completely one should watch Tyler DeWitt first and then do the hard problems with Khan.
    Very good video I understood everything.

  3. when doing stoichiometry please refrain from using fractions of an element or molecule. They can not be broken in real life, so why confuse things here.

  4. why would you say that the NO produced is 0.8 mol instead of 1 when O2 is the limiting reagent and thats the factor where only that amount of the product can be produced?

  5. I would so appreciate one of these with no fractions. Your videos have always helped me in the past, but the whole point of limiting reagents is NOT having fractions. This was kind of confusing, overall good but from a teaching standpoint PLEASE make another video with a much simpler problem. Thank you so much!

  6. in the balancing of the equation you have written 5/4 = 2.5 and then because of this error you have multiplied the whole equation by 4 instead of 2

  7. Is it okay to do it the alternate way without the fractions? I tried balancing the problem before hand amd i got the same balanced equation just did it with diffrent numbers.

  8. Listen for anyone who is confused by the fractions you shouldnt becuase you should already know how to balence the equasions and if you dont then you should go back and learn before you attemp this

  9. 15 minutes!!! for 1 question are u kidding me…
    u are going a little too much into detail…u must cut to the chase as soon as u can! I mean it's like u are teaching ABCD…Z all over again…

  10. can't get the last part where the limiting reagent , i get why if there are 2 moles of NH3 there are 2.5 moles of O2.
    If there are 1 mole O2 , there are 0.8 moles of NH3 , how and where did you get the 0.8???

  11. To balance equation, do the Oxygen last (which will make it 2.5 on the left), and then simply multiply the whole equation by 2.

  12. I completely followed along. I watched 21 straight videos from about 3-4 days ago. This series is awesome b/c they are all in order. I really am going to be at least somewhat prepared for my college chemistry course.

  13. I think so you have explained this very well. There is no amount of complication. You need to write it down and think over every step he is explaining.

  14. I cant understand how to predict the product except combustion cause it's standard… How come N+O2(g) = NO(g)
    Why not N2O??
    Im definitely missing some rule or something here.

  15. Way too much waffleing (just chatter) ,makes this harder to follow than several other similar videos. Dude, work on that.. like, I don't need to hear the term l'limiting reagent' twice when he writes it, and I don't need to hear the commentary about using different colors or making sure the number doesn't look like a 10. Just do that stuff silently. It's a pity.

  16. "not to be confused with Nitrous which is laughing gas, but i dont want to divert too much"… proceeds to talk about cigarette fumes

  17. this is the worst video Khan has produced. He is usually great, but he led me into a tantrum over this BS. not only did he make it confusing af with balancing the equation, but he didnt explain how tf O2 is the limiting reagent in terms of the RATIO between O2 and NH3. kinda just pulled it out of his booty hole

  18. I think it's great how some random YouTube video can teach me this concept better and more thoroughly in less that 20 minutes than my Chemistry professor can in an hour

  19. This is the video my Chemistry teacher had us watch, I'm currently failing that class and these equations seem too difficult for us High School students.

  20. In the first problem, (btw sorry if I didnt write the complete equation am too lazy) but instead of multplying H20 with 1.5 and all of the complicated decimal problems occured…. can't we just write it H30? Is it restricted to do that?

  21. You balanced the equation in the worst way possible … much better to make odd even and then work on the single element O2

  22. I'm teaching myself chemistry. I really like these Khan video's. I understand everything so far but I'd like a work book to practice these myself anyone have any good suggestions?

  23. I can't believe people are complaining how in depth this guy is going in explaining this topic. It is needed m8 cuz this topic is complicated

  24. Thank you so much!! I was not knowing what to do so I decided to watch your video because I haven't seen nothing about Stoquiometry, so I feel much better now with this video. I wasn't understanding the mole and stuff and thank you so much!!

  25. U made it quite easy….can you make complete lectures series on thermodynamics subjects on each chapter of it?

  26. Can someone please explain why at 10:30 he started dividing all the numbers by 5? Is it because of Oxygen being the limited reactant in the equation?

  27. Hey guys for those who got a little confused with the balancing of the equation start with the Hydogen and look for the Lowest common denominator between the hydrogen from both sides of the equation.
    Unbalanced equation:
    NH3+O2=NO+H2O
    N 1×4 | N 1×4
    H 3×4 | H 2×6
    O 2×5 | O (1×4)+(1×6)

    Balanced equation:
    4NH3+5O2=4NO+6H2O

    The lowest common denominator between 2 and 3 is 6 :
    2 4 6 8 10 {12}
    3 6 9 {12}
    If you do this with 6 you will get stuck with the incorrect number of oxygen,so rather use 12 which is the second lowest denominator.
    Hope that helped <3

  28. But then a side reaction would be the oxygen with the NO to produce nitrogen dioxide gas in which if you dissolve it into water you can get nitric acid then you can dissolve silver metal. In which you will not get the perfect numbers meaning you will have less of that ammonia reacting.

  29. This was a horrible video. Definitely didn't connect the dots like he usually does on this one. Just confusing. Not going to be able to do this skill on different equations on my own.

  30. W8 I'm a little confused here..How come that NH3 has 2 moles while O2 has only one mole. You see there are two Oxygen since one oxygen contains 16g per mole and there are a total of 32 oxy, then that means there are two moles of oxygen instead of one.? Why? Also what makes NH3 two moles… Pls help guys, Im a bit confused …need answers pls….

  31. he made balancing the equation way more complicated than it needed to be, it's just about finding lowest common denominator, no fractions otherwise i love his vids

  32. Why is this comment section full of negative thoughts
    What he did in the first 4-5 minutes was a basic pre-requisite for learning stoichiometry
    you should also know how to balance the equation using fractions
    just try to equate two elements on both sides and then use fractions to balance the the third element…………Isn't this easy??
    Why is everyone complaining and blaming his method of teaching if you cannot understand such basic thing!!

  33. the best lecture on stoichiometry on youtube ever seen is of gopal jha prosight academic hub you wil get all your doubt clear and that to in hindi

  34. Anyone else find it too chatty? Like not straight to the point I kept getting distracted and having to rewind, good video/explanation but you could cut the time in half tbf

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