Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

Question

2Al(s) + 3Cl2(g) -> 2AlCl3(s)

You are given 19.0g of aluminum and 24.0g of chlorine gas.

a)If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0g of aluminum?
= 0.704mol AlCl3

b)If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0g of chlorine gas, Cl2?
=0.255mol AlCl3

By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces the lesser amount of product.

what exactly do I do to solve for part "c"?????

c)What is the maximum mass of aluminum chloride that can be formed when reacting 19.0g of aluminum with 24.0g of chlorine?
answer:_____g AlCl3

is it?: 0.704mol AlCl3-0.255molAlCl3=0.449mol AlCl3 x 226.6g/mol = 119.7g AlCl3?

Answer 1

I tried this answer & it was incorrect.

Answer 2

First, I don't get 0.255 for mols AlCl3 formed from 24.0 g Cl2.
I get 24/71 x (2 mols AlCl3/3 mols Cl2) = 0.338 x 2/3 = 0.225 mols AlCl3.
Check my arithmetic.

Then the answer to part c is simple. Notice that the problem says, and I quote, "Keep in mind that the limiting reactant is the one that produces the lesser amount of product." So which produces the lesser amount of product? Of course that is the 24 g Cl2 so 0.225 mols AlCl3 (which is smaller than 0.704) is the # mols and that times the molar mass AlCl3 will give you the mass AlCl3. I get 0.225 x 133.3 = 30.03 g which is rounded to 30.0 to three significant figures. Check my work closely. I think 30.0 is the answer to part c but you try it, too.

Answer 3

ok...so I shouldn't have multiplied through by 2. That makes a lot more sense. Thank you, yet again, and it came out alright.

Answer 4

(oh...slightly dislexic...yes...I meant 0.225 moles, that would have helped if I had noticed before & typed it in correctly.)

Answer 5

Chemlast champer vacancies

Answer 6

how do you do part a?