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### Topic: Calculating Theoretical Yield  (Read 11724 times)

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#### wizkiddrummer

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##### Calculating Theoretical Yield
« on: May 27, 2005, 01:08:40 AM »
???I can't figure out how to find the theoretical mass and moles of Al with the following:

2Al + 3CuCl2 YIELDS 3Cu + 2AlCl3

I had .496 grams of CuCl2 to begin with. And the CuCl2 is the limiting.

I tried:

.496/134.451*(3/2)

-or-

(.496 grams CuCl2) * (1 mol/134.451) * (3 mol CuCl2/2 mol AlCl3) which equals .00553 which is what I got theoretical moles.

I then multiplied that by the molar mass of Al: 26.981539 and I got .149 grams for the theoretical moles of Al.

I guess I'm just not sure if you can find the theoretical moles of AlCl3 and then just multiply it by the molar mass of Al and have that work out.
« Last Edit: June 03, 2005, 07:28:12 AM by geodome »

#### xiankai

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##### Re:VERY URGENT ? FOR FINAL - Theoretical Yield
« Reply #1 on: May 27, 2005, 01:59:57 AM »
there's something wrong in your calculations:

.496/134.451*(3/2)

.496/134.451 is the number of CuCl2 moles, right? why do u divide it by 2? there are 3 CuCl2 moles, not 2.

i dont think u can find the theoretical mass and moles of total Al, only the mass and moles that react with CuCl. thats because Al is in excess.

"I guess I'm just not sure if you can find the theoretical moles of AlCl3 and then just multiply it by the molar mass of Al and have that work out."

that will work!, its the same as your above equation, except that u did the wrong thing in finding out 2 moles. only u'll find the mass of Al that reacts with the CuCl and not that excess total, of course.
one learns best by teaching

#### wizkiddrummer

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##### Re:VERY URGENT ? FOR FINAL - Theoretical Yield
« Reply #2 on: May 28, 2005, 01:39:30 AM »
Thank you very much for the help.  I got a 100% on the final.

-Seth

#### xiankai

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##### Re:VERY URGENT ? FOR FINAL - Theoretical Yield
« Reply #3 on: May 28, 2005, 04:35:22 AM »
heh its nothing much, only a mathematical error.