[Namossa]

Hi I have a few questions concering a lab

copper sulfate pentahydrate CuSO₄.5H₂O

1 g of copperoxid, Cuo dissolve in 10 cm^3 2M H₂SO₄ and 50 cm^3 water.
CuO(s) + 10H+ (aq) --> CuSO₄.5H2O

My question is
1. Is my formel balanced correctly?
2. Determine which reactant is limiting the exhange and why?
3. Determine the theoretical exhange? I dont know how to calculate this please help

Thank you for helping out!
Sincerely
Namossa

[Borek]

1. Is my formel balanced correctly?

Is the number of atoms of each element before and after the reaction identical? Is the total charge of all substances involved before and after the reaction identical?

[Namossa]

I think its identical it will form ions in solution but after the reaction forms CuSO₄.5H₂O

The limisting reactant is H because it needs 10 H+ in first reaction.

Any ideas how to solve the rest of the questions?

Thank you!

[Namossa]

Further update information:
the water comes from dissolve solution. Number of crystal water that bound on solid compound is dependent on temperature. In high temperature exist waterfree phases. In the solution might have sulfuric acid which dont disturb the continuation process.

[Babcock_Hall]

You have a sulfur atom in the product that is not on the reactant side.  The equation must be balanced properly before you can perform a limiting reagent calculation.

[Namossa]

My misstake i did not write is down
however here is the balanced formula:

CuO(s) + 10H+(aq) SO₄^2- --> CuSO₄.5H2O

Can you please calculate the limiting reagent calculation?

[Borek]

I can only repeat what I wrote earlier:

Count oxygen atoms on the left and on the right. Are their numbers identical?

What is the net charge on the left? On the right? Are they identical?