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August 2, 2015

Homework Help: Chemistry

Posted by Jeremy on Tuesday, September 28, 2010 at 11:19pm.

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!

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