CBSE Class 11 > Class 11 Videos > L15 : Limiting Reagent - Basic Concepts of Chemistry, Class 11
L15 Limiting Reagent - Basic Concepts of Chemistry, Class 11 Video Lecture
FAQs on L15 : Limiting Reagent - Basic Concepts of Chemistry, Class 11
| 1. What is a limiting reagent in chemistry? |  |
Ans. In chemistry, a limiting reagent is a reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed. It determines the maximum amount of product that can be obtained.
| 2. How do you determine the limiting reagent in a chemical reaction? | |
Ans. To determine the limiting reagent in a chemical reaction, you need to compare the given amounts of reactants and their stoichiometric coefficients. The reactant that produces the least amount of product, based on the balanced chemical equation, is the limiting reagent.
| 3. What happens if there is an excess of a reactant in a chemical reaction? | |
Ans. If there is an excess of a reactant in a chemical reaction, it means that there is more of that reactant than required for the reaction to reach completion. The excess reactant will not be completely consumed and will be left over after the reaction is finished.
| 4. Can the same reactant be the limiting reagent in different reactions? | |
Ans. Yes, the same reactant can be the limiting reagent in different reactions. The concept of limiting reagent depends on the stoichiometry of the reaction and the given amounts of reactants. If the stoichiometric ratios and reactant amounts align in a way that one reactant is limited, it can be the limiting reagent in multiple reactions.
| 5. How does the limiting reagent affect the yield of a reaction? | |
Ans. The limiting reagent directly affects the yield of a reaction. The amount of product formed is determined by the limiting reagent. If there is an insufficient amount of the limiting reagent, the reaction will not proceed to completion, resulting in a lower yield. Conversely, if there is excess of one reactant, the yield will still be limited by the amount of the limiting reagent present.
Text Transcript from Video
hello friends this video on basic
concepts of chemistry part 15 is brought
to you by example a comm normal for you
from exam before watching this video
please make sure that you have watched
part 1 on working now understand
something called limiting reagent in a
chemical reaction sometimes we see that
you have reactant one reactant
it gives product so if let's suppose
this guy is less examples in this
reaction we know that 10 grams of this
reacts with 20 grams of this to give
product you Mickey these 20 but instead
of 10 you make it 5 what will happen
after 5 after this guy is exhausted this
will not be able to react this guy will
act as a limiting reagent because this
guy is lessening quantity then required
so for example in this typical reaction
10 grams for this one and 20 grand for
this was required to form some product
since you have 10 grams this is only to
5 grams so after all this 5 grams is
consumed this is not there so reaction
is stopped right so this guy is limiting
the reaction and that's why it's called
limiting H that means you can see that
in a chemical reaction the reactant
which is present in lesser amount then
it is required right the one which is
within glycerin some assumes it get
consumed after some time and no reaction
take place even if you keep adding these
guy the reaction won't happen right
because this guy is not there thus that
particular reactant is limiting reagent
because that guy limits the amount of
reaction you put more of these reactions
will get started because this guy has
less in quantity right so this guy is
all over now I didn't even keep pumping
this nothing will happen because this
guy needs the action one reactant one to
the act but there is no reactant one
only the action to the reaction won't
happen right so in the reaction the the
reactant which is present in lesser
amount then required that is called
integration please note it is not for
example I tell you there is a reaction
where
20 grams of reactant 1 and let's suppose
5 grams of reactant to was required
right my Shero 20 let's suppose I have
10 grams of AF in 1 plus 5 and really I
can do in this case also this guy is my
limiting reagent plus it but if you see
this guy is 10 this guy is 5 so don't
just compare these two numbers see you
you can't say that in this guy
this reaction this guy is staying then
phase 5 grams with this guy will be
leave it again it is not the case the
case is for 5 grams are reacting to 20
grams of the Eternals pad but in this
case I have only 2 grams so you have to
just find how many grams of reactant one
how many grams of reactant to are
required then you have to find if you
find the ratio which one is lesser so in
this case for 5 grams of reactant to my
20 grams of reactant along scrap forget
this case I have only 10 grams of
reaction 1 so this guy is my limiting
reagent though 10 is more than 5 and but
this guy is my limiting reagent so don't
just compare the numbers of all the
reactant and say the one this is less as
a related limiting reagent that is not
the way the correct way is you have to
find a chemical reaction we need to find
how many grams of this how many grams
are required and then based on that you
defined that for example in this case
this guy is X this guy is 20 this 4 X
multiplied outside so it is fine
this has to be 20 this less so this is a
remedy
we'll take some examples of limiting
reagent so we have 5 kg of nitrogen 10
kg of hydrogen are combined to form
ammonia you have to find the ammonia gas
form and you to find the limiting
reagent in this situation so if you see
my reaction is something like this my n2
plus h 2 gives
nh3 let's balance this this is n 2 to
ammonia to nitrogen or liquid to six
hydrogen but
six there's a balanced equation right so
this I have 50 kg and this I have but
let's see how much it is required okay
let's find that then from this reaction
I can say that one mole of n2 requires
three mole of h2 correct so one mole of
n2 is 20 grams plus 2 28 grams of into
requires 3 2 hydrogen h2 is 2 that is 6
grams of s T so I can say that 28 grams
of n2 required 6 gram of h2 all right so
one gram of hydrogen requires how many
grams of nitrogen 29 by 6 grams of
nitrogen so I have 10 kg of a hydrogen
so 10 kg of hydrogen will require how
much 28 by 6 do they don't Wilbur 3
correct this many grams because I am in
grams only now that one but 46.67 kg a
minute kitchen a correct so my 10 grams
of this requires 46 kg but I have 50 kg
so if you see this way is extra this guy
is extra but I am looking for limiting
reagent which is less that means this
guy is this see in this case I found
that 10 grams of hydrogen acquired 46 kg
so this 10 grams of hydrogen will react
46.67 kg will be gone but remaining will
have some three point are 3 3 kg left my
Trojan and there will be no hydrogen
left so hydrogen gets consumed early so
hydrogen is my limiting reagent I hope
you understand this part because in this
case I found that one gram of hydrogen
8:26 8x6 name of nitrogen and 10kg of
hydrogen needs 46.67 kg of nitrogen but
we have 50 kg of nitrogen that means
once this once the 10 grams of hydrogen
is consumed this many grams of nitrogen
gazoo and remitting is left so my
impression is left hydrogen gets
consumed early so hydrogen ism limiting
reagent now I got my limiting reagent
that becomes my critical thing not now I
have to find the first question is job
done identify the limiting region we
have to find calculate ammonia in form
now since my hydrogen is limiting
reagent I have to rely on the hydrogen I
can't rely on nitrogen because whole of
the nitrogen is not consumed correct I
should rely on hydrogen so I can see
from this reaction three mole of
hydrogen the mole of hydrogen gives two
mole of ammonia one of Aniston or 3 into
hydrogen is to 6 grams of hydrogen lives
do we do this one how many more is the
molar mass of ammonia ammonia will be
nitrogen n is our 14 last 3 17 into the
30 42 into 17 that is 34 grams 1 so 6
grams of hydrogen gives 34 grams of
ammonia so I can say that my 10 kg of
hydrogen will give how much 34 was 6
into 10 kg of ammonia on that is 56.66
so that is miles please note in this
case I am relying only on hydrogen
because hydrogen is my limiting reagent
so I say that my 3 moles of hydrogen
gets 2 moles of ammonia that is 6 times
of hydrogen gives 44 grams of ammonia
that is 10 kg of hydrogen gives 56.66
p0m correct thank you
visit exam viacom to watch free
education videos try free online test
get the best quality study materials
study from the best tutors and mentors
and much more
thank sense again
concepts of chemistry part 15 is brought
to you by example a comm normal for you
from exam before watching this video
please make sure that you have watched
part 1 on working now understand
something called limiting reagent in a
chemical reaction sometimes we see that
you have reactant one reactant
it gives product so if let's suppose
this guy is less examples in this
reaction we know that 10 grams of this
reacts with 20 grams of this to give
product you Mickey these 20 but instead
of 10 you make it 5 what will happen
after 5 after this guy is exhausted this
will not be able to react this guy will
act as a limiting reagent because this
guy is lessening quantity then required
so for example in this typical reaction
10 grams for this one and 20 grand for
this was required to form some product
since you have 10 grams this is only to
5 grams so after all this 5 grams is
consumed this is not there so reaction
is stopped right so this guy is limiting
the reaction and that's why it's called
limiting H that means you can see that
in a chemical reaction the reactant
which is present in lesser amount then
it is required right the one which is
within glycerin some assumes it get
consumed after some time and no reaction
take place even if you keep adding these
guy the reaction won't happen right
because this guy is not there thus that
particular reactant is limiting reagent
because that guy limits the amount of
reaction you put more of these reactions
will get started because this guy has
less in quantity right so this guy is
all over now I didn't even keep pumping
this nothing will happen because this
guy needs the action one reactant one to
the act but there is no reactant one
only the action to the reaction won't
happen right so in the reaction the the
reactant which is present in lesser
amount then required that is called
integration please note it is not for
example I tell you there is a reaction
where
20 grams of reactant 1 and let's suppose
5 grams of reactant to was required
right my Shero 20 let's suppose I have
10 grams of AF in 1 plus 5 and really I
can do in this case also this guy is my
limiting reagent plus it but if you see
this guy is 10 this guy is 5 so don't
just compare these two numbers see you
you can't say that in this guy
this reaction this guy is staying then
phase 5 grams with this guy will be
leave it again it is not the case the
case is for 5 grams are reacting to 20
grams of the Eternals pad but in this
case I have only 2 grams so you have to
just find how many grams of reactant one
how many grams of reactant to are
required then you have to find if you
find the ratio which one is lesser so in
this case for 5 grams of reactant to my
20 grams of reactant along scrap forget
this case I have only 10 grams of
reaction 1 so this guy is my limiting
reagent though 10 is more than 5 and but
this guy is my limiting reagent so don't
just compare the numbers of all the
reactant and say the one this is less as
a related limiting reagent that is not
the way the correct way is you have to
find a chemical reaction we need to find
how many grams of this how many grams
are required and then based on that you
defined that for example in this case
this guy is X this guy is 20 this 4 X
multiplied outside so it is fine
this has to be 20 this less so this is a
remedy
we'll take some examples of limiting
reagent so we have 5 kg of nitrogen 10
kg of hydrogen are combined to form
ammonia you have to find the ammonia gas
form and you to find the limiting
reagent in this situation so if you see
my reaction is something like this my n2
plus h 2 gives
nh3 let's balance this this is n 2 to
ammonia to nitrogen or liquid to six
hydrogen but
six there's a balanced equation right so
this I have 50 kg and this I have but
let's see how much it is required okay
let's find that then from this reaction
I can say that one mole of n2 requires
three mole of h2 correct so one mole of
n2 is 20 grams plus 2 28 grams of into
requires 3 2 hydrogen h2 is 2 that is 6
grams of s T so I can say that 28 grams
of n2 required 6 gram of h2 all right so
one gram of hydrogen requires how many
grams of nitrogen 29 by 6 grams of
nitrogen so I have 10 kg of a hydrogen
so 10 kg of hydrogen will require how
much 28 by 6 do they don't Wilbur 3
correct this many grams because I am in
grams only now that one but 46.67 kg a
minute kitchen a correct so my 10 grams
of this requires 46 kg but I have 50 kg
so if you see this way is extra this guy
is extra but I am looking for limiting
reagent which is less that means this
guy is this see in this case I found
that 10 grams of hydrogen acquired 46 kg
so this 10 grams of hydrogen will react
46.67 kg will be gone but remaining will
have some three point are 3 3 kg left my
Trojan and there will be no hydrogen
left so hydrogen gets consumed early so
hydrogen is my limiting reagent I hope
you understand this part because in this
case I found that one gram of hydrogen
8:26 8x6 name of nitrogen and 10kg of
hydrogen needs 46.67 kg of nitrogen but
we have 50 kg of nitrogen that means
once this once the 10 grams of hydrogen
is consumed this many grams of nitrogen
gazoo and remitting is left so my
impression is left hydrogen gets
consumed early so hydrogen ism limiting
reagent now I got my limiting reagent
that becomes my critical thing not now I
have to find the first question is job
done identify the limiting region we
have to find calculate ammonia in form
now since my hydrogen is limiting
reagent I have to rely on the hydrogen I
can't rely on nitrogen because whole of
the nitrogen is not consumed correct I
should rely on hydrogen so I can see
from this reaction three mole of
hydrogen the mole of hydrogen gives two
mole of ammonia one of Aniston or 3 into
hydrogen is to 6 grams of hydrogen lives
do we do this one how many more is the
molar mass of ammonia ammonia will be
nitrogen n is our 14 last 3 17 into the
30 42 into 17 that is 34 grams 1 so 6
grams of hydrogen gives 34 grams of
ammonia so I can say that my 10 kg of
hydrogen will give how much 34 was 6
into 10 kg of ammonia on that is 56.66
so that is miles please note in this
case I am relying only on hydrogen
because hydrogen is my limiting reagent
so I say that my 3 moles of hydrogen
gets 2 moles of ammonia that is 6 times
of hydrogen gives 44 grams of ammonia
that is 10 kg of hydrogen gives 56.66
p0m correct thank you
visit exam viacom to watch free
education videos try free online test
get the best quality study materials
study from the best tutors and mentors
and much more
thank sense again
About this Video
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Apr 20, 2026 Last updated
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