Question

If 42.5 g N2 react with 10.1 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over?

N2 3H2 ---> 2NH3

Answer 1

@Frostbite

Answer 2

That should be N2 + 3H2 ---> 2NH3

Answer 3

Alrightie.
In this question we shall be using what we used last time and find the limiting reactant. We assume that the reaction will run to the end, so that the reaction will first stop when one of the reactants has been depleted.
How would you attack the problem if you had to guess? :)

Answer 4

Well I know the limiting reactant is going to be H2

Answer 5

And you know that because? :)

Answer 6

Because H2 can only run 3 times vs N2 being able to run 22 times

Answer 7

Oh, wait, N2 can run 42 times

Answer 8

Well, there is a fast little note I wanna point out. That kind of trick we did last time only works when we talk of molecules (amount of substance), you are talking about the masses of the substances. And as we know the mass for each molecule is different.

Answer 9

So first we need to calculate the amount of substance \(n\).

Answer 10

You know how to calculate the amount of substance?

Answer 11

I'm not really sure :/

Answer 12

Okay, the formula is:
\[\Large n=\frac{ m }{ M }\]
Here is \(n\) the amount of substance, \(m\) the mass and \(M\) the molar mass. The molar mass, is the sum of the individual atoms in the molecule.
The amount of substance \(n\) is nothing but the number of molecules in the solution of the specific compound in the unit "mole".

Answer 13

The reasons why this is important is because we read your chemical equation like this:
N2 + 3H2 ---> 2NH3
1 mole of molecular nitrogen + 3 mole of molecular hydrogen reacts to given 2 mole of ammonia

Answer 14

Ok, so for N2 I would do 42.5/28?

Answer 15

Yes and remember units :) (teachers love units)

Answer 16

Ok :)

Answer 17

\(\approx\)1.52

Answer 18

mol

Answer 19

Alright so we got 1.52 mol of \(\sf N_{2}\). do the same with hydrogen :)

Answer 20

10.1 g/2.01588 g/mol

Answer 21

\(\approx\)5.01

Answer 22

Alright perfect. So we got
\(n(N_{2})=1.52~ mol\)
\(n(H_{2})=5.01~ mol\)
So as we can see we got must molecules of hydrogen, but the ratio of reactant use is 3:1, how would you take this into consideration when finding the limiting reactant?

Answer 23

Know it is a little tedious work, but if you get the hang of it, you can do these questions like with a snap of your fingers.

Answer 24

Do it like last time?

Answer 25

Yes you could use that. :) but use the general way :)

Answer 26

Or just to give a reminder:
We devide \(n(H_2)\) with 3 to check.
So we get:
1.67 mol times would the reaction run if \(\sf H_{2}\) was the limiting reactant.
1.52 mol times would the reaction run if \(\sf N_{2}\) was the limiting reactant.
So which one is it? :)

Answer 27

H2 is the limiting reactant

Answer 28

Hmmm which one got the must molecules? :)

Answer 29

Even if we divide hydrogen with 3 to take into consideration it uses 3 molecules compared to H2 that only use 1.

Answer 30

1.67/3 and 1.52/1?

Answer 31

I've already divided with 3, sorry should let you do that your self:
5.01/3 & 1.52/1, I think I might confused you a little, sorry.

Answer 32

oh

Answer 33

But what do you get then? :)

Answer 34

1.67 and 1.52

Answer 35

Exactly. so what is the limiting reactant?

Answer 36

N2?

Answer 37

Exactly! :)

Answer 38

Now from here it is all calculations :)

Answer 39

But first, do you understand what you have done so far and why?

Answer 40

Yes :D

Answer 41

Perfect, any clue what we do from here? :)

Answer 42

ummmm
no :(

Answer 43

Okay. Now that we know we that nitrogen is the limiting reactant, we use that information to calculate the amount of substance for all the other chemicals.
For example: we know that hydrogen is being used so we say:
\[\Large n_2(H_2)=n(H_2)-3 \times n(N_2)\]
The subscript 2 means the reaction has run to the end. The reason for this equation is that the amount of substance left for hydrogen, must be the total amount before the reaction \(n(H_2)\) minus what has been used, which is equal to \(3 \times n(N_2)\)
Does that make since?

Answer 44

yes

Answer 45

Good. try calculate AFTER you answer me this question:
What is the amount of substance for nitrogen when the reaction is done \((n_2(N_2))\)?

Answer 46

1.67-3*1.52
1.67-4.56
-2.89?

Answer 47

Sorry, \(n(H_2)=5.01\) the number 1.67 was simply to figure out the amount of "moles" the reaction could be ran, and can be forgotten at this point.

Answer 48

Oh ok

Answer 49

So 5.01-4.56
.45

Answer 50

Exactly that is \(\large n_2(H_2)\). :)
What is \(\large n_2(N_2)\)? Think on this one.

Answer 51

5.01?

Answer 52

Naaa think :)
\(N_2\) was the limiting reactant... and what was special for the limiting reactant? (hint: when does the reaction stop?)

Answer 53

Are you still here? :)

Answer 54

yes

Answer 55

Use it with caution :)

Answer 56

This document sum up our discussions here.

Answer 57

Wow! That is awesome! Thank you so much

Answer 58

Wanna continue trying here while having the document? :)

Answer 59

Or think you will do?

Answer 60

I think I will continue here

Answer 61

Altight so where were we, oh yes the amount of molecules of \(N_2\) when the reaction is over?

Answer 62

0?

Answer 63

Exactly? and why? :)

Answer 64

I think you: cause it is the limiting reactant, the reaction stop when the limiting reactant got no molecules left.

Answer 65

I think you know*

Answer 66

Ok, that makes sense

Answer 67

It is chemistry, it is all logic :)
Anyway. What about ammonia? Try write an equation that describe the amount of substance for ammonia like I for hydrogen.

Answer 68

I don't think I understand what you are saying :/

Answer 69

Try calculate the amount of substance for ammonia: \(n_{2}(NH_3)\)

Answer 70

And explain how you do it :)

Answer 71

2*N2
2*1.54
3.02

Answer 72

And why do we multiple with two? (hint: look into the reaction)

Answer 73

to produce 2 moles of ammonia?

Answer 74

Perfect! :D
So now we got the amount of substance for all of them.
Try calculate the mass. Explain what you do with text during the calculations

Answer 75

remember units :)

Answer 76

(If you can explain it to me, I know you understand it and I can sleep well tonight)

Answer 77

haha
I'm not really sure how to explain other than to multiply the moles by the grams/mol
.45 mol * 2 g/mol = .9 g H2
3.04 mol * 17.03 g/mol = 51.8

Answer 78

Well.... the units go out so it is logical to multiple them together right?
Also it is just another variant of the equation: n = m/M <-> m = n * M