February 27, 2017

Homework Help: CHEM-last part prob.

Posted by K on Tuesday, October 16, 2007 at 12:22am.

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s) + 3Cl2(g) -> 2AlCl3(s)

You are given 19.0g of aluminum and 24.0g of chlorine gas.

a)If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0g of aluminum?
= 0.704mol AlCl3

b)If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0g of chlorine gas, Cl2?
=0.255mol AlCl3

By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces the lesser amount of product.

what exactly do I do to solve for part "c"?????

c)What is the maximum mass of aluminum chloride that can be formed when reacting 19.0g of aluminum with 24.0g of chlorine?
answer:_____g AlCl3

is it?: 0.704mol AlCl3-0.255molAlCl3=0.449mol AlCl3 x 226.6g/mol = 119.7g AlCl3?

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