How many liters of a .500M #H_2SO_4# must react with excess #Cu# to produce 15.0 g #CuSO_4#?

1 Answer
Jul 12, 2018

Answer:

As @ernest-z and @anor277 have pointed out, a dilute solution of #"H"_2 "SO"_4# is unable to oxidize #"Cu"# under standard conditions.

This answer demonstrates the calculation of the number of moles of #ul("concentrated")# #"H"_2 "SO"_4 # necessary to produce the same amount of #"Cu" "SO"_4# from excess #"Cu"# metal.

Explanation:

Copper is too stable to react with dilute aqueous solutions of sulfuric acid (such as this one of concentration #0.500 color(white)(l) "mol" * "dm"^(-3)#) under standard conditions.

It takes concentrated sulfuric acid (which is indeed oxidizing but only at high concentrations) to directly react with copper and produce #"CuSO"_4#. Some orange-colored #"SO"_2(g)# gas evolves in this process. The reaction takes place as follows:

#stackrel(color(navy)(0))("Cu")(s) + 2 color(white)(l) "H"_2 stackrel(color(purple)(+6))("S") "O"_4("conc." ,aq ) #
#to stackrel(color(navy)(+2))("Cu") stackrel(color(grey)(+6))("S") "O"_4 (aq) + stackrel(color(purple)(+4))("S")"O"_2(g) + "H"_2"O"(l)#

This equation can be balanced by tracking changes in the oxidation state of #"Cu"# and #"S"#. From the reaction equation:

#(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = color(purple)(2)/color(blue)(1)#

#n(color(blue)("CuSO"_4)) = (m(color(blue)("CuSO"_4)))/(M(color(blue)("CuSO"_4))) = (15.0 color(white)(l) "g")/ (159.6 color(white)(l) "g" * "mol"^(-1)) = 0.0940 color(white)(l) "mol"#

#n(color(purple)("H"_2"SO"_4)) = n(color(blue)("CuSO"_4)) *(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = 0.0940 color(white)(l) "mol" * 2 = 0.188 color(white)(l) "mol"#

Meaning that it takes #0.188 color(white)(l) "mol"# of (concentrated) sulfuric acid to produce #15.0 color(white)(l) "g"# of #"CuSO"_4# through direct oxidation of copper metal #"Cu"#.