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Chemistry

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Hi, I posted my question earlier but I was having some difficulty with it.
2Al+ 3Cl2-->2AlCl3
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:

Cl2-36.0g/70.9= 0.507mol
Al-31.0g/26.98=1.15mol

Therefore Chlorine is the limiting reagent, since I have less that 3/2.

Cl2:AlCl3= 3:2
0.507mol (3/2)=0.76 mol AlCl3.

Please check my answer and significant figures. I am pretty sure I messed up!

  • Chemistry -

    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.

  • Chemistry -

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  • Chemistry -

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