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

Jul 12, 2018

As @ernest-z and @anor277 have pointed out, a dilute solution of ${\text{H"_2 "SO}}_{4}$ is unable to oxidize $\text{Cu}$ under standard conditions.

This answer demonstrates the calculation of the number of moles of $\underline{\text{concentrated}}$ ${\text{H"_2 "SO}}_{4}$ necessary to produce the same amount of ${\text{Cu" "SO}}_{4}$ from excess $\text{Cu}$ metal.

#### Explanation:

Copper is too stable to react with dilute aqueous solutions of sulfuric acid (such as this one of concentration $0.500 \textcolor{w h i t e}{l} {\text{mol" * "dm}}^{- 3}$) under standard conditions.

It takes concentrated sulfuric acid (which is indeed oxidizing but only at high concentrations) to directly react with copper and produce ${\text{CuSO}}_{4}$. Some orange-colored ${\text{SO}}_{2} \left(g\right)$ gas evolves in this process. The reaction takes place as follows:

$\stackrel{\textcolor{n a v y}{0}}{\text{Cu")(s) + 2 color(white)(l) "H"_2 stackrel(color(purple)(+6))("S") "O"_4("conc.} , a q}$
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 $\text{Cu}$ and $\text{S}$. From the reaction equation:

$\left(n \left(\textcolor{p u r p \le}{{\text{H"_2"SO"_4))) /(n(color(blue)("CuSO}}_{4}}\right)\right) = \frac{\textcolor{p u r p \le}{2}}{\textcolor{b l u e}{1}}$

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 $0.188 \textcolor{w h i t e}{l} \text{mol}$ of (concentrated) sulfuric acid to produce $15.0 \textcolor{w h i t e}{l} \text{g}$ of ${\text{CuSO}}_{4}$ through direct oxidation of copper metal $\text{Cu}$.