posted by Jeremy on .
Hi, I posted my question earlier but I was having some difficulty with it.
You are given 31.0g of aluminum and 36.0g of chlorine gas.
1.If you had excess chlorine, how many moles of aluminum chloride could be produced from 31.0g of aluminum?
This is my work so far:
Therefore Chlorine is the limiting reagent, since I have less that 3/2.
0.507mol (3/2)=0.76 mol AlCl3.
Please check my answer and significant figures. I am pretty sure I messed up!
Yes, you did.
The first part is not a limiting reagent since you have all of the Cl2 needed.
moles Al = 31.0/26.98 = 1.149
moles AlCl3 = 1.149 moles Al x (2 moles AlCl3/2 moles Al) = 1.149 x (1/1) = 1.149 moles AlCl3 which rounds to 1.15 to three s.f.
#2a You can do it two ways. The way I suggested, since you had to do #1 anyway, is to keep #1 in mind and work #1 with 36.0 g Cl2 and all the Al needed.
moles Cl2 in 36.0 g is
36.0/70.9 = 0.50776 (which I would have rounded to 0.508 BUT I usually carry extra places(in the calculator) and round at the end of the problem.
moles AlCl3 = 0.50776 mols Cl2 x (2 moles AlCl3/3 moles Cl2) = 0.3385 moles which rounds to 0.338 moles AlCl3. (Your error is in the conversion factor.) Then I compare moles from #1 where we had all of the Cl2 needed (1.15) with #2 where we had all of the Al we needed (0.338) and the correct value is the smaller of the two; that is 0.338 moles AlCl3 formed.
2b. The other way Bob Pursley showed you is the way you did it to determine Cl2 was the limiting reagent.
31.0/26.98 = 1.15 moles Al and
(31.0/26.98) x (3 moles Cl2/2 moles Al) = 1.72 moles Cl2 needed and we don't have that much.
(36.0/70.9 = 0.508 moles Cl2.
(36.0/70.9) x (2 moles Al/3 moles Cl2) = 0.338 moles Al needed and we have that much; therefore, your work is ok in determining the limiting reagent.
You gotta take the bleeble blabble and add it to sazzle shnazzle.
You must find the value for x and then twirk on a poster of Miley Cyrus for an hour.