Homework Help: Chemistry

Posted by Mercedes on Tuesday, September 1, 2009 at 12:12am.

A chemist is producing copper(II) chloride using the following reaction:
Cu(s) + Cl2(g) --> CuCl2(s)
In one reaction, the chemist uses 15.0 g of copper metal and 10.0 g of chlorine gas as reactants. When the reaction reaches completion, 16.0 g of copper(II) chloride has been produced. What was the percent yield of the reaction?

Chemistry - Jim_R, Tuesday, September 1, 2009 at 2:02am
must be each reactant ?

or total reactants ?

Ill do each

Step one:

mm of products
cu=63.546
2cl=70.90
------------
134. 45g/mol

step two
mol of products

16.0 g/134.45=1.19 x 10^-1 mol x

step two

find the lr

cu=15.0 /63.546 =.236mol produced

cl=10.0/70.90=.141

lr is cl

three

turn it into grams

.141mol x 134. 45g/mol=18.96 g

four % yield

16.0g/18.96 g x 100=84.4% around.

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