# A #"1.85 g"# sample of mixture of #"CuCl"_2# and #"CuBr"_2# was dissolved in water and mixed thoroughly with a #"1.8 g"# portion of #"AgCl"#. After reaction, the solid, a mixture of #"AgCl"# and #"AgBr"#, was filtered, washed, and dried, and [ . . . ] ?

##
[ . . . ] its mass was found to be #"2.052 g"# . What percent by mass of the original mixture was #"CuBr"_2# ?

[ . . . ] its mass was found to be

##### 1 Answer

**DISCLAIMER:** *COMPLICATED ANSWER!*

Let's write down what we know.

#m_"start" = m_(CuCl_2) + m_(CuBr_2) = "1.85 g"# #" "" "bb((A))#

#m_"end" = m_(AgCl) + m_(AgBr) = "2.052 g"# #" "" "bb((B))#

#2"AgCl"(s) + "CuBr"_2(aq) -> 2"AgBr"(s) + "CuCl"_2(aq)#

Notice how **the only source of** **is from the starting**

#1.8 cancel"g AgCl" xx cancel"1 mol AgCl"/(143.32 cancel"g AgCl") xx "1 mol Ag"^(+)/cancel"1 mol AgCl"#

#=# #"0.01256 mols Ag"^(+) = "0.01256 mols starting AgCl"#

As a sidenote, we should check that the

#0.01256 cancel"mols starting AgCl" xx cancel"1 mol AgBr"/cancel"1 mol AgCl" xx "187.77 g AgBr"/cancel"1 mol AgBr"#

#=# #"2.358 g AgBr dry yield"# , which is more than the dry yield of both solids combined of#"2.052 g"# . So,#"AgCl"# is in excess.

Next, it is also useful to note that **the only source of** **on the reactants side is from the**

#n_(Br^(-)) = (m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2)# #" "" "bb((1))#

*Now our goal is to find expressions for the mass of each component in the produced precipitate.*

Assume we know what mols of

#n_(Br^(-)) = n_(AgBr)# #" "" "bb((2))#

Knowing the mols of

Since

#"0.01256 mols AgCl reactant" - n_(AgBr)#

#= n_("leftover AgCl")# #" "" "bb((3))#

Its mass would then be given by:

#m_("leftover AgCl") = "143.32 g/mol" xx n_("leftover AgCl")# #" "" "bb((4))#

We're getting there. Now, we need an expression for the mass of

#=> m_(AgBr) = overbrace((m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2))^(n_(AgBr)) xx "187.77 g"/"mol"#

Now we have expressions for the mass of leftover

#m_("leftover AgCl") + m_(AgBr) = "2.052 g"#

#= "143.32 g/mol" xx n_("leftover AgCl") + "187.77 g"/"mol" xx n_(AgBr)#

- For mols of leftover
#"AgCl"# , after plugging#(1)# and#(2)# into#(3)# , plug#(3)# into our current form of#(B)# . - For mols of
#"AgBr"# , plug in#(1)# and#(2)# .

#=> "143.32 g/mol" xx ["0.01256 mols AgCl reactant" - (m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2)] + "187.77 g"/"mol" xx (m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2)#

To make this easier to read, let

#= {143.32 xx [0.01256 - 0.008954x]} "g" + 1.6812x " g"#

#= {1.8001 - 1.2833x + 1.6812x} " g"#

#= {1.8001 + 0.3980x} " g" = "2.052 g"#

#=> x = 0.2519/0.3980 " g" = "0.6329 g CuBr"_2#

Finally, we have the mass! We can use this mass along with the mass given from **percent by mass**:

#color(blue)(%"w/w CuBr"_2 ) = "0.6329 g CuBr"_2/"1.85 g copper mixture" xx 100%#

#~~ color(blue)(34.21%)#