Question

Help

Answer 1

Ehem alright I haven't done this in a while but let's start with calculating the number of moles of N2 and Mg we have. Can you do that?

Answer 2

It doesn't tell you which is the excess or limiting, so I guess we'll have to solve for that.

You need two ratios to determine this

1. (problem ratio) 8 moles of mg/2 moles of n2
2. (equation ratio) 3 moles of mg/1 mole of n2

problem>equation

so numerator is excess, denominator is limited

mg=excess
n2=limited

Now..

Take the problem limited (2) in this case, multiply it by the excess of the equation (excess must be on numerator) (3).

That will form 6 moles of Mg3N2.

Get the molar mass of Mg3N2 and multiply it by 6.

Answer 3

@Jhannybean
Check please.

Answer 4

I've done a few problems like these, all of them confuse the hell out of me, and forces me to look back at my notes everytime...

Let me try and see if I did anything wrong...I'm sure I'm close

Answer 5

so then it would be

100.9494 x 6, if i'm right. I wish someone would check for me.

Answer 6

I know for sure that before we multiply the 6 by the molar mass.

the 6 represents that the limiting reactant (2 moles of n2) will all be used, while the excess (mg), only 6 moles will be used.

Answer 7

What do you have so far?

Answer 8

Actually, I have a feeling you ARE right.

If your equation of

3MG+N2>>MG3N2

And there is actually 6 moles of MG being used, and 2 moles of nitrogen..

Wouldn't that just mean we multiply the whole equation by...2?

2Mg3n2=202g

Answer 9

Why do I feel as if I'm missing something? What happened to the multiplying of 6? I'm all out of coffee so my brain at this point has shut down.

Answer 10

Is there answer choices by any chance?

Answer 11

Hmm, let me look over my regents book real quick and see if I can get any info.

Answer 12

I'm just stuck at that final part.

Here's an example in my regents book.

How many grams of water can be made according to the equation:

2H2+o2>>2h20?

If 100 grams of h2 and 160 grams of o2 are bought together to react?

with all that ratio mumbo jumbo..

h2=excess
o2=limiting

then they did exactly what I did.

take the limiting of the problem(5), multiply it by the excess of the equation (2)

and the finally the molar mass of h20.

they got 180 grams of h20.

Answer 13

So unless, I'm missing something which I'm trying so hard to see (but I can't), my first answer was correct.

Answer 14

Let @Jhannybean check it. It has to be one of the two answers we made.

Answer 15

\[\sf 3 Mg(s) + \color{red}{1}N_2(g) \longrightarrow Mg_3N_2(s)\] How many grams of product are formed from 2.0 mol of N\(_2\) (g) and 8.0 mol of Mg(s)?\[\sf \text{2 mol}~ N_2 \times \frac{\text{3 mol}~Mg}{\text{1 mol}~ N_2} = 3~ mol~ Mg\]\[\sf 8.0~ mol~Mg \times\frac{\text{1 mol}~ N_2}{\text{3 mol}~Mg} =2.6~ mol~ N_2\]\[\therefore \sf N_2 ~\text{is the LR}\]

Answer 16

Now, which one is the limiting reagent?

Answer 17

So you're looking for the grams of product FORMED, this can only be found by the LR, which is \(\sf N_2\). Use the amount of LR to find grams of \(\sf Mg_3N_2\) produced. \[\sf 2.6~mol~N_2 \times \frac{1~mol~ Mg_3N_2}{1~mol~N_2}\times \frac{100.9~g~Mg_3N_2}{1~mol~Mg_3N_2}=262~g~ Mg_3N_2 ~produced \]

Answer 18

There are several ways to find the LR, there is no one direct way.

Answer 19

@Jhannybean Can you briefly look over my work to see what I did wrong?

Answer 20

Awwman..... Haha jk jk one min.

Answer 21

@dtan5457 I think where you might have messed up is there is only 1 mol of N\(_2\) there, or you could say 2 moles of N.

But there is NOT 2 moles of N\(_2\) being used. that is absolutely wrong.

Answer 22

\[\sf 2~mol~N_2 \implies 2N_2\]

Answer 23

How many grams of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s)?

If you are already given the mole of one reactant, you can easily know how many it's being used by the product by finding which reactant is the limiting regent. To find the limiting reagent. YOU use mole to mole ratio to calculate the moles of the product that is being produced.

In this case, our two reactants are N2 (nitrogen gas) and Mg( magnesium solid)

\(2.0~ mol ~N2∗\Large\frac{1 mol Mg3N2}{1 mol N2}∗\Large\frac{100.9 ~g~Mg3N2}{1~mol~Mg3N2}=202 ~g~Mg_3N_2\)

\(8.0~mol~Mg∗\Large\frac{1~mol~Mg3N2}{3~mol~Mg}∗\Large\frac{100.9~g~Mg3N2}{1~mol~Mg3N2}=270~g~Mg3N2\)

Answer 24

1. (problem ratio) 8 moles of mg/2 moles of n2
2. (equation ratio) 3 moles of mg/1 mole of n2

the problem does say 2 moles of n2.

the equation is 1 mole, so I don't think that was my mistake.

Answer 25

mols of reactant----->Mole to mole ratio-----> Molar mass----> mass

Answer 26

I'm pretty sure your solving this differently than I am, but I'm learning everything from a book, and they are doing exactly what I'm doing xD

Answer 27

1. take the moles of each reactant and find how many moles of the reactant are reacting with EACHOTHER.

2.compare the amount of moles that are PRODUCED from the reactants reacting with eachother.

3. The smaller amount of moles will tell you which one is the LR.

4. Use the limiting reactant to find out how many moles..grams...etc are produced of the precipitate.

Answer 28

That's how I was taught to do limiting reactant problems.

Answer 29

Look at what my book says though

Answer 30

they are telling me to do dreadful ratios

Answer 31

Ok let me try this problem

Answer 32

okay, but try to look follow what my book says, (as that is the only way I understand it)

Answer 33

maybe the book is wrong ^_^

Answer 34

Ok, first off I see they are converting to moles \(\checkmark\)

Answer 35

which we don't have to do here for @Itsokay problem, but yeah.

Answer 36

We could, and it's easier to see which one is the LR then

Answer 37

Let me write it out so theres not so many words. Words are annoying in writing out these problems.

Answer 38

\[\sf 2H_2 +O_2 \longrightarrow 2H_2O\] How many grams of \(\sf H_2O\) can be made if 100.0 g \(\sf H_2\) and 160.0 g \(\sf O_2\) are reacted together. \[\sf 100.0 ~g~H_2 \times \frac{2~mol~H_2}{2.02~g~H_2} = 99.0~mol~H_2 \]\[\sf 160.0~g~O_2 \times \frac{1~mol~O_2}{32.00~g~O_2}=5~mol~O_2\]Compare the two and you find that \(\sf O_2\) is the LR

Answer 39

I'm pretty sure what your book shows is how @Jhannybean and @Zale101 did it

Answer 40

I mean this is the method i'm most comfortable with.

Answer 41

are you getting 180 grams as well, though?

Answer 42

Then you take the moles of \(\sf O_2\) and find how many grams were produced. \[\sf 5.000~mol~O_2 \times \frac{2~mol~H_2O}{1~mol~O_2}\times \frac{18.00~g~H_2O}{1~mol~H_2O}= 180.0~g~H_2O \]

Answer 43

I had to fix my number of sig figs, it was 5.000 instead of just 5.

Answer 44

In the original problem, using your method, would it not just be 2 moles x molar mass of product?

Answer 45

as n2 is the limiting reactant

Answer 46

Hey guys. Long chat here, heh. I will post my work.

Answer 47

i mean, you had 2.6 moles @Jhannybean

Answer 48

Yeah, you're right @dtan5457 , but you always have to show the mole to mole ratio. It's how you know you're moving from 1 compound to the other.

Answer 49

I believe the method shown in your book follows my method, so if anything is the problem there, go back to my post and see if I used the correct molar mass. then punch it into your calculator and you'll find the answer.

Answer 50

but why is it 2.6 moles though.

Answer 51

I believe my answer is absolutely correct.

Answer 52

if the question called for 2?

Answer 53

You're being asked about 8.0 moles of magnesium, but only 3 moles are reacting with the 1 mole of nitrogen.

Answer 54

Therefore \(\dfrac{8.0}{3} =2.6\)

Answer 55

@Zale101 did not show which one was the limiting reactant. She went directly from the reactants to the product, but this does not show if any,of the reactant, if at all, was completely used.

Answer 56

According to the question, it is important to take account of the molar ratio of the reactants. Mg to N2 is 3:1.
Secondly, considered the given moles, we have to decifer which of reactants is limiting and which one is in excess.
If we are given 8 mol of Mg, we expect it to react with (8*1)/3=8/3moles of N2, which is more than the 2 moles of N2 available. Meaning, N2 is the limiting reactant and hence, we work out the resulting mass of the product from this reagent.
The molar ratio of N2 to Mg3N2 is 1:1, automatically, if we have 2 mol of N2, we also have 2 mol of Mg3N2. To find the mass of reaction , we use the formula m=n*Mr, where m is the required mass, Mr is the molecular mass and n is the number of moles. Therefore, we get m=2*[(24*3)+(14*2)]=200g of Mg3N2.
That is it.

Answer 57

When you get to more advanced chemistry,. you need to know which compounds completely ionize, and which ions are reacted to produce the precipitate, therefore at the end of the reaction, there will be no more of the limiting reactant ions left in your solution (because they will be all used up).

Therefore its important to always clearly state which one is the limiting reactant, and which one is the excess.

Answer 58

another different answer xD

Answer 59

our answers are ranging from 200-600 grams

Answer 60

no one is using my method so I still don't see what I did wrong

Answer 61

It all really depends on the number of sig figs you either keep or round up in your work. Thats what differentiates the grams of precipitate produced. None of the answers are wrong, theyre all above 200 g which is good.

Answer 62

I bet my answer is correct. At least I am very confident with it.

Answer 63

yeah but i got 600 grams at first, which is pretty significant.
if you want to, look at my attachment and the way my book did it, and see what answer you come up with.

Answer 64

Let's see you solve it your books method, @Itsokay and see what you get. Compare it to our answers. I have not seen your work here so work it out and compare.

Answer 65

@Jhannybean why did you use 2.6 mol of N2 while you only have 2 moles of it available?

Answer 66

Alright, and if you happen to get the exact correct answer by your professor or something, let us know :)

Answer 67

Yay!!

Answer 68

\(\color{blue}{\text{Originally Posted by}}\) @Itsokay
She used 2.6 because in the initial reaction it is 3Mg(s), therefore 3 moles of Mg(s). I get it, now.
\(\color{blue}{\text{End of Quote}}\)

That tells me you understand how I went about solving the equation, therefore you understand how to find the LR.

Answer 69

But 3 mol of Mg are to 1 of N2
And 8 mol of Mg are to 8/3 mol of N2.
Conversely, you are provided with only 2 mol of N2, meaning there are 2.6-2=0.6mol of N2 not available, which are out the 2 available.

Answer 70

@Jhannybean

Answer 71

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[\sf 3 Mg(s) + \color{red}{1}N_2(g) \longrightarrow Mg_3N_2(s)\] How many grams of product are formed from 2.0 mol of N\(_2\) (g) and 8.0 mol of Mg(s)?\[\sf \text{2 mol}~ N_2 \times \frac{\text{3 mol}~Mg}{\text{1 mol}~ N_2} = 3~ mol~ Mg\]\[\sf 8.0~ mol~Mg \times\frac{\text{1 mol}~ N_2}{\text{3 mol}~Mg} =2.6~ mol~ N_2\]\[\therefore \sf N_2 ~\text{is the LR}\]
\(\color{blue}{\text{End of Quote}}\)

You're told you are using 2.0 moles of N\(_2\), that does not mean mean you subtract it by the limiting reactant. Conversely, the 2.0 moles is not a definite amount. The reaction equation tells definitely that you will need only 1 mole of nitrogen reacted with 3 moles of magnesium to produce the compound Magnesium Nitride.

Also, the only "available" amount comes from the limiting reactant, which you find from the reaction equation. THAT is what tells you how much of each is reacted, and how much is available.

Answer 72

The 2.0 moles of N\(_2\) and 8.00 moles of Mg are just amounts you are using. they can also account for error.

Answer 73

I have never seen someone work on the same problem for 3 hours straight and still question it.

I want to amend your dedication here.

Answer 74

my first answer was 600 grams so I don't think I would know LOL

Answer 75

Sorry,you caught my typo :) it's 6.

Answer 76

\[\sf 2 \cancel{\text{ mol}~ N_2} \times \frac{3\text{ mol}~Mg}{1\cancel{\text{ mol}~ N_2}} = 2(3\text{mol Mg}) = 6~ mol~ Mg\]

Answer 77

It doesn't affect anything, the LR is still N\(_2\). STOP OVER THINKING THIS!!!!! Hahaha

Answer 78

Ok follow these steps, you'll be fine after this:

1. given grams or moles of reactants, convert BOTH reactants to moles.

2. Use the moles of one to find the moles of the other

3. compare how many moles of each reactant is produced

4. use the smaller one as your LR to find how much of the product you are asked to find.

Answer 79

never seen a question go on this long before lol. maybe we broke the record ?

Answer 80

"Jhannybean: @Zale101 did not show which one was the limiting reactant. She went directly from the reactants to the product, but this does not show if any,of the reactant, if at all, was completely used"

I didn't show the limiting reactant, but i showed the process of how to solve it.
I dont solve everything but leave some for the tutee to do on their own and that is why I asked "What was the limiting reactant?"

Itsokay said it was N2 which is correct. I see no point in tutoring and solving everything without doing the process step by step and not getting the tutee to participate.
~That is just my style

Another reason was, @Jhannybean tutored and explained everything clearly afterwards in which i then didn't see no point of me to do further explanations ^_^

Thanks @itsokay for participating :)

Answer 81

Great Job everyone!

Answer 82

Oh ok, I see now :) Good work @Zale101 :)