Posted by K on Tuesday, October 16, 2007 at 12:22am.
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
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?
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?
is it?: 0.704mol AlCl3-0.255molAlCl3=0.449mol AlCl3 x 226.6g/mol = 119.7g AlCl3?
- CHEM-last part prob. - K, Tuesday, October 16, 2007 at 12:24am
I tried this answer & it was incorrect.
- CHEM-last part prob. - DrBob222, Tuesday, October 16, 2007 at 12:34am
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.
- CHEM-last part prob. - K, Tuesday, October 16, 2007 at 12:43am
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.
- CHEM-last part prob. - K, Tuesday, October 16, 2007 at 1:21am
(oh...slightly dislexic...yes...I meant 0.225 moles, that would have helped if I had noticed before & typed it in correctly.)
- CHEM-last part prob. - Maikalraj, Wednesday, June 5, 2013 at 2:30am
Chemlast champer vacancies
- CHEM-last part prob. - ashley, Monday, October 21, 2013 at 4:38pm
how do you do part a?
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