Question

Please help:)

Answer 1

@Vincent-Lyon.Fr @bronzegoddess Please help:)

Answer 2

Do you want solution or hints?

Answer 3

only answers becoz i have done it. no need of solutions or hints.

Answer 4

i want to check only.

Answer 5

a. work out how many grams of ammonia each reactant produces, the one that gives the least is the limiting reagent.
i gtg, but i'll look at it again :)

Answer 6

a) Nitrogen is the limiting reagent
b) 1214.3 g of ammonia are produced
c) excess of hydrogen is 35.7 g

You can check your solution by conservation of mass as:
1000 + 250 = 1214.3 + 35.7

Answer 7

k.
thanx a lot @Vincent-Lyon.Fr sir.
I did right.

Answer 8

Well done Mahesh! You're the one who deserves the medal!

Answer 9

thanx again sir:)

Answer 10

Now, if I may, I can ask a follow up question:
If yield is only 80%, what are the final masses of reactants and product?
This might help you fully understand this kind of problems.

Answer 11

sorry to say but i haven't studied %yield yet
bt i will give the answer after upto 4 hrs:)

Answer 12

100% i will give sir:)

Answer 13

np ;-)

Answer 14

:)

Answer 15

@Vincent-Lyon.Fr , I studied %yield but not mastered it btw i thought the answer of the question u asked me: -
I have thought that Reactants will be (1000+250) = 1250gm. & product will be
\[1250 \times \frac{80}{100}=1000gm.\]Am i right or nt sir?

Answer 16

No, you will end up with 80% ammonia of the value you had before.

Answer 17

so i took the question in wrong way
i thought that u r asking me a new question
but it was a follow-up:(

Answer 18

ok sir one more question
would the reactant's mass would be the same as before?
I think yes!

Answer 19

No because if yield is not 100%, then the excess of reactant will be greater.

Answer 20

sir would u plz make me understand the concept of %yield?
I think i have nt gained it yet:(

Answer 21

A certain amount of reactants is used up; a certain amount of product is formed.

If yield is n%, then multiply these amounts by n% and this is it.

Answer 22

oh i see but
sir i think that %yield must not have effect on reactants becoz it is yielding the products!

Answer 23

wt do u say sir?
Plz help:)

Answer 24

@Vincent-Lyon.Fr sir plz:)

Answer 25

Instead of 35.7 moles N2 being used up, only 28.57 will be used up.

200 g N2 and 78.6 g H2 will remain, and 971.4 g of NH3 will be formed.

Answer 26

k; sir.
Thnx a lot for removing my nonsense confusion.

Answer 27

sorry to bother but how did you in fact solve the questions?

Answer 28

@bronzegoddess ? Pardon me! wt r u saying? i didn't understand.

Answer 29

how did you find N2 to be the limiting reagent?

Answer 30

becoz \[N_2.\]is that reactant which will be used up before the reaction goes to completion:)

Answer 31

I think u want solution; do u?

Answer 32

yes

Answer 33

k Plz give me the mole ratio of reactants & products from the balanced reaction
then we'll go further.

Answer 34

4moles:2moles

Answer 35

ya but in actually i wanted
1:3:2.
u gave me
1+3:2

Answer 36

okay sorry, i thought you want all reactants to all products..

Answer 37

no no
k now we move further

Answer 38

calculate the moles of H2 & N2 from the given mass & tell me:)

Answer 39

okay, for N2 and H2; 35.7moles and 125moles respectively.

Answer 40

k; if 1 mole of N2 gives 3 moles of H2
therefore 35.7 moles of N2 will produce 107.1 moles
but we have actually 125 moles of H2
so H2 is in excess & limiting reagent would be N2 which limits the reaction
if we had some more N2 then total 125 moles would be used up:)
then there will be no limiting reagent. all would be balanced.
So now any doubt?

Answer 41

nope, thank you! :)

Answer 42

^_^