A chemist is producing copper(II) chloride using the following reaction:

Question

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?

Answer 1

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.