February 23, 2017

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

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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