Class 11 Exam > Class 11 Videos > L16 : Limiting Reagent Numerical - Basic Concepts of Chemistry, Class 11
L16 : Limiting Reagent Numerical - Basic Concepts of Chemistry, Class 11 Video Lecture
FAQs on L16 : Limiting Reagent Numerical - Basic Concepts of Chemistry, Class 11 Video Lecture
| 1. What is a limiting reagent in chemistry? |  |
Ans. A limiting reagent is a reactant that is completely consumed during a chemical reaction, thereby limiting the amount of product that can be formed. It determines the maximum amount of product that can be obtained.
| 2. How do you identify the limiting reagent in a chemical reaction? | |
Ans. To identify the limiting reagent, you need to compare the mole ratio of the reactants to the stoichiometric ratio in the balanced chemical equation. The reactant with the smallest mole ratio compared to the stoichiometric ratio 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 is required to react with the limiting reagent. The excess reactant will not be completely consumed and will be left over after the reaction is complete.
| 4. Can a reactant be both a limiting reagent and an excess reagent in the same reaction? | |
Ans. No, a reactant cannot be both a limiting reagent and an excess reagent in the same reaction. The limiting reagent is the reactant that is completely consumed, while the excess reagent is the reactant that is left over. They are mutually exclusive.
| 5. How does the concept of limiting reagent relate to the yield of a chemical reaction? | |
Ans. The concept of limiting reagent is closely related to the yield of a chemical reaction. The amount of product obtained in a reaction is limited by the amount of the limiting reagent. If there is an excess of one reactant, it does not contribute to the yield of the reaction. Theoretical yield is the maximum amount of product that can be obtained based on the limiting reagent, while actual yield is the amount of product actually obtained in the reaction.
Text Transcript from Video
hello friends this video on basic
concepts of chemistry part 16 is brought
to you by exam fluid Tom no more for you
from exam before watching this video
please make sure that you have watched
part 1 part 15 the question says need to
find the amount of carbon dioxide
produced when one mole of carbon is
going to no sizzle so see the reaction
the reaction is seedless o2 is equal to
CO 2 from this reaction say that one
mole of carbon needs 1 mole of oxygen to
give 1 mole of carbon dioxide so the
intention carbon carbon dioxide also
only so we can see that one mole of
carbon gives 1 mole of C so first answer
the question is all one mole of carbon
how much amount of carbon dioxide is
formed when the mole of carbon is one
answer is 1 mole of C second is one mole
of carbon is burned in 16 grams of wood
so you see from this reaction one mole
of carbon is 1 mole of C 2 that is we
can see that 12 gram of carbon gives 1
mole of C 2 is 12 plus 16 into 32 32
plus 12 is 44 right 12 plus 2 into 16
carbon is processed there is 44 grams of
sieved
so first I can say one mole of co2 a
particular programs of co2 is mines for
a second I know that my 12 grams of
carbon it needs how much car gram of
oxygen let's find this because I have I
have only 60 grams of dioxin I am told
that one mole of carbon that is one mole
of carbon plus my one mole of co2 gives
one mole of CO correct this final
equation one mole of carbon is 12 gram
of carbon C needs one mole of oxygen is
32 grams of oxygen give one mole of co2
I told 44
see you now since my this guy is only 16
grams not 32 gram so this oxygen will
act as a limiting reagent correct since
oxygen is acting as a limiting reagent I
know that 32 grams of oxygen gives 44
grams of carbon dioxide so 16 grams of
oxygen will yield 44 by 32 into 16 that
is nothing but 22 grams of this answer
is also incurred also there are two
moles of carbon burn with 16 grams of by
oxygen here also you see the oxygen is
less because it was one mole hit you are
adding more power now but oxygen is less
right oxygen is the limiting reagent so
now things will depend only on oxygen
hey loss will get the same answer
because here also if you see since
oxygen is limiting origin 16 grams of
oxygen will always form from you grams
of carbon dioxide so here also you get
the savings hope you understand because
here also if you see oxygen is my
limiting region there is a reaction a
plus B 2 is equal to a B 2 you define
the limiting reagent in this reaction
where I am told that three hundred atoms
of a and two hundred molecules of B see
here a plus B 2 is equal to AV D here I
can say that one atom one atom of a
since it is atom reacts with one
molecule of B molecule of B to give a
little okay in fact one molecule of you
please note this is Adam this is
molecule so my one atom of this reacting
with one molecule of a B to give a veto
so
I have my 200 molecule here because they
have 200 molecules of a so I have 200
molecules of B if you make that lead how
much 200 times of me please note if one
molecule of B needs one atom of a so 200
molecules are be willing to advance of
it so you see they have 300 atoms over
here so a is extra since a is extra my B
is limiting reagent hope you understand
this in this case two molecules of B so
two and molecules of B needs to n atoms
of a so after that all B is consumed we
have extra a since a is extra my B is
limiting regime in case of B I have two
molecules of a and three modules are be
reacting so if I have two molecules of a
and three molecules of B so if I have
two molecules or atoms you can say is
atoms also you can say is atomic way so
two atom of a will need two molecule of
D so two molecule of be required but we
have only three extra we have three
modulo B so we have extra V molecule
here so here my a is limiting reagent
correct so how do you hope you
understand this in this case my B is
limiting reagent and in this case a
limiting you why because in first case
we see that all my B molecules are
getting consumed a is extra in this case
I am seeing that all my a molecules are
getting as you my B's extra right so the
one which get consumed early is called
limiting reagents in in first case might
be my limiting reagent
in second East is my limiting reagent
because there are two molecules away so
that will react with two molecules of B
so you'll see one molecules of B is
extra so B is mine sorry a is my
limiting reagent in this case the
question says dinitrogen and by hydrogen
reacts with each other to produce
ammonia as per this reaction
nitrogenous hydrogen gives to nh3 we can
balance this so this is 3 1 this is
balanced on you tomorrow find the mass
of ammonia produced if 2 e to the power
3 grams of by nitrogen reacts with one
gram some by Harrison in this we can see
that the grams of might by nitrogen and
by hydrogen is given that means 1 may be
a limiting reagent and that's where the
question says will any of this remain
reactant remain unreacted if yes what
would be the mass so busy with this I
can say that one mole of nitrogen we add
for 3 mole of hydrogen 2 gives 2 mole of
ammonia right I can write like this 1
mole of nitrogen reacts with 3 mode off
of h2 gives to Moloch in s3 one mole of
nitrogen how much 20 grams you can see
then 20 grams of nitrogen reacts with 3
moles of hydrogen is 3 into 2 6 grams of
hydrogen okay
hi this is 34 MLAs 34 to make this Y
will be 7 7 plus 7 plus law sorry 14
plus 3 17 into 244 this 34 grams aw now
you see we are told that this this is
our it's 2 kg of nitrogen reacts with 1
kg of a hydrogen whether he actually get
itself we can see that 20 in grams of
nitrogen required six gram of hydrogen
so hydrogen require is less right
nitrogen require is more so that is we
can see that just looking at this only
we can see that my intrusion will be a
limiting reagent in this right because
nitrogen requires pretty more actually
you see right and hydrogen is less
again we can see this because 28 grams
required for six grams right but here it
is only two brands for one capsule that
but we can try to exactly find how many
grams of net required source
so let's assume that 28 times of
nitrogen I am right here 28 grams of
into requires 6 grams of h2 so one gram
of n2 will require 6 by 28 grams of h2
and I have 2 kg of nitrogen or 2 into 10
to the power 3 grams of n2 will require
6 by 28 into to the power 3 grams and
the scum of the wheel 4.28 6 into a 2 by
3 grams that is 4 to the 6 grams for
twenty eight point six times this is
what will come out this ramps off
hydrogen is required how many hands of
hydrogen we have we have thousand grams
of hydrogen so we have hydrogen extra
right so hydrogen will remain unreacted
and my limiting limiting reagent is
nitrogen here because nitrogen is less
correct my I need only for 26 grams of
for 28.6 Rams of hydrogen I have 1000
grams of hydrogen so hydrogen is extra
right so I my hydrogen left will be I
have to find how many grams of hydrogen
remains unreacted so hydrogen unreacted
is equal to total hydrogen present
initially - hydrogen got reactive
correct that will be 571 5 volts will be
left this will be unreacted this part is
done
and that's what my answer is also I have
to find the mass of ammonia produced
since my nitrogen's the limiting reagent
I will work on my person so I am told
that my 20 grams of nitrogen gives 34
grams of ammonia so my 28 grams of
nitrogen implies 44 grams of ammonia I
am taking nitrogen because my original
aromatic legend so in this case act 2
gives you of nitrogen so 2 into 10 to
the power 3 grams of n2 will give you 34
by 28 into 2 into 10 to the power 3
grams and that will come out to be 240
8.57 where these many grams of ammonia
thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more
thanks and Celine
concepts of chemistry part 16 is brought
to you by exam fluid Tom no more for you
from exam before watching this video
please make sure that you have watched
part 1 part 15 the question says need to
find the amount of carbon dioxide
produced when one mole of carbon is
going to no sizzle so see the reaction
the reaction is seedless o2 is equal to
CO 2 from this reaction say that one
mole of carbon needs 1 mole of oxygen to
give 1 mole of carbon dioxide so the
intention carbon carbon dioxide also
only so we can see that one mole of
carbon gives 1 mole of C so first answer
the question is all one mole of carbon
how much amount of carbon dioxide is
formed when the mole of carbon is one
answer is 1 mole of C second is one mole
of carbon is burned in 16 grams of wood
so you see from this reaction one mole
of carbon is 1 mole of C 2 that is we
can see that 12 gram of carbon gives 1
mole of C 2 is 12 plus 16 into 32 32
plus 12 is 44 right 12 plus 2 into 16
carbon is processed there is 44 grams of
sieved
so first I can say one mole of co2 a
particular programs of co2 is mines for
a second I know that my 12 grams of
carbon it needs how much car gram of
oxygen let's find this because I have I
have only 60 grams of dioxin I am told
that one mole of carbon that is one mole
of carbon plus my one mole of co2 gives
one mole of CO correct this final
equation one mole of carbon is 12 gram
of carbon C needs one mole of oxygen is
32 grams of oxygen give one mole of co2
I told 44
see you now since my this guy is only 16
grams not 32 gram so this oxygen will
act as a limiting reagent correct since
oxygen is acting as a limiting reagent I
know that 32 grams of oxygen gives 44
grams of carbon dioxide so 16 grams of
oxygen will yield 44 by 32 into 16 that
is nothing but 22 grams of this answer
is also incurred also there are two
moles of carbon burn with 16 grams of by
oxygen here also you see the oxygen is
less because it was one mole hit you are
adding more power now but oxygen is less
right oxygen is the limiting reagent so
now things will depend only on oxygen
hey loss will get the same answer
because here also if you see since
oxygen is limiting origin 16 grams of
oxygen will always form from you grams
of carbon dioxide so here also you get
the savings hope you understand because
here also if you see oxygen is my
limiting region there is a reaction a
plus B 2 is equal to a B 2 you define
the limiting reagent in this reaction
where I am told that three hundred atoms
of a and two hundred molecules of B see
here a plus B 2 is equal to AV D here I
can say that one atom one atom of a
since it is atom reacts with one
molecule of B molecule of B to give a
little okay in fact one molecule of you
please note this is Adam this is
molecule so my one atom of this reacting
with one molecule of a B to give a veto
so
I have my 200 molecule here because they
have 200 molecules of a so I have 200
molecules of B if you make that lead how
much 200 times of me please note if one
molecule of B needs one atom of a so 200
molecules are be willing to advance of
it so you see they have 300 atoms over
here so a is extra since a is extra my B
is limiting reagent hope you understand
this in this case two molecules of B so
two and molecules of B needs to n atoms
of a so after that all B is consumed we
have extra a since a is extra my B is
limiting regime in case of B I have two
molecules of a and three modules are be
reacting so if I have two molecules of a
and three molecules of B so if I have
two molecules or atoms you can say is
atoms also you can say is atomic way so
two atom of a will need two molecule of
D so two molecule of be required but we
have only three extra we have three
modulo B so we have extra V molecule
here so here my a is limiting reagent
correct so how do you hope you
understand this in this case my B is
limiting reagent and in this case a
limiting you why because in first case
we see that all my B molecules are
getting consumed a is extra in this case
I am seeing that all my a molecules are
getting as you my B's extra right so the
one which get consumed early is called
limiting reagents in in first case might
be my limiting reagent
in second East is my limiting reagent
because there are two molecules away so
that will react with two molecules of B
so you'll see one molecules of B is
extra so B is mine sorry a is my
limiting reagent in this case the
question says dinitrogen and by hydrogen
reacts with each other to produce
ammonia as per this reaction
nitrogenous hydrogen gives to nh3 we can
balance this so this is 3 1 this is
balanced on you tomorrow find the mass
of ammonia produced if 2 e to the power
3 grams of by nitrogen reacts with one
gram some by Harrison in this we can see
that the grams of might by nitrogen and
by hydrogen is given that means 1 may be
a limiting reagent and that's where the
question says will any of this remain
reactant remain unreacted if yes what
would be the mass so busy with this I
can say that one mole of nitrogen we add
for 3 mole of hydrogen 2 gives 2 mole of
ammonia right I can write like this 1
mole of nitrogen reacts with 3 mode off
of h2 gives to Moloch in s3 one mole of
nitrogen how much 20 grams you can see
then 20 grams of nitrogen reacts with 3
moles of hydrogen is 3 into 2 6 grams of
hydrogen okay
hi this is 34 MLAs 34 to make this Y
will be 7 7 plus 7 plus law sorry 14
plus 3 17 into 244 this 34 grams aw now
you see we are told that this this is
our it's 2 kg of nitrogen reacts with 1
kg of a hydrogen whether he actually get
itself we can see that 20 in grams of
nitrogen required six gram of hydrogen
so hydrogen require is less right
nitrogen require is more so that is we
can see that just looking at this only
we can see that my intrusion will be a
limiting reagent in this right because
nitrogen requires pretty more actually
you see right and hydrogen is less
again we can see this because 28 grams
required for six grams right but here it
is only two brands for one capsule that
but we can try to exactly find how many
grams of net required source
so let's assume that 28 times of
nitrogen I am right here 28 grams of
into requires 6 grams of h2 so one gram
of n2 will require 6 by 28 grams of h2
and I have 2 kg of nitrogen or 2 into 10
to the power 3 grams of n2 will require
6 by 28 into to the power 3 grams and
the scum of the wheel 4.28 6 into a 2 by
3 grams that is 4 to the 6 grams for
twenty eight point six times this is
what will come out this ramps off
hydrogen is required how many hands of
hydrogen we have we have thousand grams
of hydrogen so we have hydrogen extra
right so hydrogen will remain unreacted
and my limiting limiting reagent is
nitrogen here because nitrogen is less
correct my I need only for 26 grams of
for 28.6 Rams of hydrogen I have 1000
grams of hydrogen so hydrogen is extra
right so I my hydrogen left will be I
have to find how many grams of hydrogen
remains unreacted so hydrogen unreacted
is equal to total hydrogen present
initially - hydrogen got reactive
correct that will be 571 5 volts will be
left this will be unreacted this part is
done
and that's what my answer is also I have
to find the mass of ammonia produced
since my nitrogen's the limiting reagent
I will work on my person so I am told
that my 20 grams of nitrogen gives 34
grams of ammonia so my 28 grams of
nitrogen implies 44 grams of ammonia I
am taking nitrogen because my original
aromatic legend so in this case act 2
gives you of nitrogen so 2 into 10 to
the power 3 grams of n2 will give you 34
by 28 into 2 into 10 to the power 3
grams and that will come out to be 240
8.57 where these many grams of ammonia
thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more
thanks and Celine
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