An alloy contains copper and zinc. Some of the zinc has become oxidized to zinc oxide. What is the result of adding an excess of dilute sulfuric acid to the alloy?
1. A blue solution and a white solid remains.
2. A colorless solution and a pink / brown solid remains.
3. The alloy dissolves completely to give a blue solution.
4. The alloy dissolves completely to give a colorless solution.
The reactivity of copper is much less than that of zinc. While zinc can react with dilute sulfuric acid copper does not react with dilute sulfuric acid, it only reacts with concentrated sulfuric acid.
The alloy is made up of copper and zinc; some of the zinc has oxidized to zinc oxide. Zinc reacts with dilute sulfuric acid following the chemical reaction: Zn + H2SO4 --> ZnSO4 + H2. Zinc oxide reacts with dilute sulfuric acid following the chemical reaction ZnO + H2SO4 --> ZnSO4 + H2O.
ZnSO4 is a colorless compound that dissolves in water. Copper settles to the bottom as a pink/brown solid.
When the alloy made of copper and zinc is reacted with dilute sulfuric acid the result is a colorless solution and a pink/brown solid lying at the bottom of the container.
I know that zinc trots into solution in sulfuric acid, and I have seen a copper wire sit for days in a beaker of sulfuric acid without any apparent change, so I'd guess that the answer in post #1 is correct.
However, if the alloy were not completely covered by the acid, air might mix with the acid resulting in some of the copper disolving, and that would cause the acid to turn blue.
Or at least that is what I think. I have not tried it with copper-zinc alloy, but I did watch a half-submerged copper wire become two wires as the copper disolved at the boundary between acid and air.