Question

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

Answer 1

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"`.