Question

(LIMING REACTANT!) Can somebody show me how to do this? I keep getting different answers every time I try it, I'm sure it's not as hard as I make it...

Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine.
2Al + 3Cl2 -> 2AlCl3
If you begin with 2.70 g of Al and 4.05 g of Cl2,
a) Which reactant is limiting?
b)What mass of AlCl3 can be produced?
c)What mass of the excess reactant remains when the reaction is completed?

Answer 1

There are two ways to solve this kind of problem. Some schools or countries teach to work with masses (usually in grams), others with amounts (in moles).

I can show you how to do it, but tell me what method is used by your teacher.

Answer 2

My teacher taught two different ways, I think he had preferred moles.

Answer 3

Ok, now what you have to to is convert all masses in you question into moles of matter.

To do this, you need to know the molar masses of all compounds in the equation.

Using the period table or any other table, complete the following:

M(Al) = ..... g/mol
M(Cl2) = ..... g/mol
M(AlCl3) = ..... g/mol

Answer 4

So far I think I got this.. I feel like I might be missing a step or something?
mass Al = 2.70g *( 1mol Al/26.982gAl) = .1000667
mass Cl2 = 4.05g*(1molCl2/70.906gCl2) = .057118

Answer 5

Correct, except that it should read (my correction in capital):

AMOUNT Al = 2.70g *( 1mol Al/26.982gAl) = .1000667 MOL
AMOUNT Cl2 = 4.05g*(1molCl2/70.906gCl2) = .057118 MOL

Answer 6

Oh okay right. So to find the amount of AlCl3, is this correct?

(2.70+4.05)*(1molAlCl3/133.341gAlCl3) = 133.341 mol AlCl3

Answer 7

No, amount of AlCl3 is not obtained this way at all.
Try to follow what I am typing.

Now build a table in moles and imagine Al is the limiting reactant:

2 Al + 3 Cl2 ---> 2 AlCl3
\(n_o\) 0.1 0.057 0
\(\Delta n\) -0.1 ?
_____________________________________________________
\(n_f\) 0.0 ?

\(n_o\) is initial amount (in moles), \(\Delta n\) is change in amount due to the reaction, \(n_f\) is final amount

Now copy the table and fill it in.

Answer 8

I don't recognize this format. Which information do I try to find first?

Answer 9

You have to find first the change in the amount of Cl2, i.e. the amount of Cl2 used in the reaction.

Answer 10

I'm still a bit lost. How do I do that?

Answer 11

The numbers in line \(\Delta n\) must be proportional to the stoichiometric numbers in the equation.

Answer 12

Let me try. Does it start like something like this?
2 Al + 3 Cl2 ---> 2 AlCl3
no 0.1 0.057 0
Δn 2(-0.1) 3(-0.1) 2(-0.1)
_____________________________________________________
nf 0.0 ?

Answer 13

No!

If Al is the limiting reactant, all initial amount (0.1 mol) will be used in the reaction (0.1 mol).

You cannot consider to have 0.1 mole at the beginning and use 0.2 mole of it. Tha's impossible. The -0.1 must stay as it is.

Answer 14

Besides, the \(\Delta n\) on the left bear a minus sign because reactants are used.
But the \(\Delta n\) on the right-hand side must bear a + sign because products are formed.

Answer 15

Would you multiply it by 2/3 then? I was thinking of the stoichiometric ratio.
2 Al + 3 Cl2 ---> 2 AlCl3
no 0.1 0.057 0
Δn -0.1 2/3(-0.1)
_____________________________________________________
nf 0.0 ?

Answer 16

No, you must multiply it by 3/2 because there are 3 Cl2 used every time 2 Al are used.

Answer 17

Okay that makes sense. I'm also trying to wrap my head around the concept. What is the next step?

Answer 18

Simply subtract amount of Cl2 used from initial amount : that would lead you to the excess of Cl2, if any.

Answer 19

Okay.. 0.057- ((3/2)*-0.1) = -.093 .. Wait, I got an answer that doesn't make sense, what did I do wrong here?

Answer 20

Great!

Now, as we have a contradiction, it means that the initial assumption (Al is limiting reactant) is false.

This proves that Cl2 is the limiting reactant.

Now build a new table in which Cl2 is used up.

Answer 21

Okay nice. Let me try..
2 Al + 3 Cl2 ---> 2 AlCl3
no 0.1 0.057 0
Δn (2/3)-0.057 -0.057
_____________________________________________________
nf 0.0 ?

Answer 22

Correct, now fill in the last line (final amounts) and the last column for AlCl3 and you are almost finished :-)

Answer 23

How does this look?
2 Al + 3 Cl2 ---> 2 AlCl3
no 0.1 0.057 0
Δn (2/3)-0.057 -0.057 +0.057
_____________________________________________________
nf 0.138 0.0 0.057

Answer 24

2 slight mistakes :

For Al : 0.1 moles at the beginning MINUS 0.038 mole used = 0.062 mol at the end

For AlCl3 : as the stoichiometric number is 2, there are also 0.038 mole formed, not 0.057 mole.

Answer 25

Okay how about this?
2 Al + 3 Cl2 ---> 2 AlCl3
no 0.1 0.057 0
Δn (2/3)-0.057 -0.057 +(2/3)0.057
_____________________________________________________
nf 0.062 0.0 0.038

Answer 26

Perfect, now convert back final line into masses. All the answers to the questions in your problem will be solved.

Answer 27

Okay.
0.062mol Al*(26.982g Al/1mol Al) = 1.673g Al

0.038mol AlCl3*(133.341g AlCl3/ 1mol AlCl3) = 5.067g AlCl3

Answer 28

Fine. Now you know how to do it :))

In order to practise, you can do it again (now or later) with the same equation, starting with:
13.5 g Al and 88.75 g Cl2

You should find: 35.5 g excess Cl2 and 66.75 g AlCl3 formed.

Answer 29

With time, you will find that it is usually easy to know which reactant is limiting, and so avoid going through the table to find out the hypothesis was the wrong one.

Answer 30

Wow okay that was really cool. Thanks for your input and thanks for your time :D

Answer 31

You're welcome :)

By the way, some people call that kind of table ICE-table (for Initial-Change-Equilibrium).

You will learn about equilibria later, so in your case, it is more an ICF (Final) table as your reaction is complete.

Answer 32

Okay that's exciting. Again that was superb, so thanks for help me practice. C: