# An other stoiciometry problem

1. Feb 15, 2005

### danne89

Hi again! The problem goes like this:
How many g CuSO4*5H_2O does one get when 20 g of a 98% H2SO4 solution and copper interact, if only 85% of the cristalls is possible to use.

$$Cu+H_2SO_4 + 5H_2O \rightarrow CuSO_4 * 5H_2O + H_2$$
20*0.98=19.6
$$\frac{19.6}{H_2SO_4} = \frac{x}{CuSO_4*5H_2O}$$
x=49.87
49.87 * 0.85 = 42.39

2. Feb 15, 2005

### dextercioby

I'm sorry,but the reaction is REDOX AND IT IS NOT CORRECT IN THE FORM U HAVE GIVEN.Please adjust it.

Daniel.

3. Feb 15, 2005

### danne89

I'm sorry too; for being so stupid.
$$Cu \rightarrow Cu^{2+} + 2e^-$$
$$H_2SO_4 + 2e^- \rightarrow H_2 + SO_4^{2-}$$
$$SO_4^{2-} + Cu^{2+} \rightarrow CuSO_4$$
$$CuSO_4 + 5H_2O \rightarrow CuSO_4 + 5 H_2O$$
Hmm. That cannot be right. Cu is a preciouser metal then H...

4. Feb 15, 2005

### dextercioby

That's right.From your reaction,there will not come out any Hydrogen.Take another chance.

Daniel.

5. Feb 15, 2005

### danne89

$$Cu + 5 H_2SO_4 \rightarrow CuSO_4 * 5H_2O +5 SO_4^{2-}$$

Edit: No! It's wrong too. Bluaa

Last edited: Feb 15, 2005
6. Feb 15, 2005

### dextercioby

No.It's wrong.If i tell u the reaction products,can i trust you'll do the redox thing...?

Daniel.

P.S.Copper(II) sulphate,sulphur dioxyde and plain water.

7. Feb 16, 2005

### danne89

Nooo. I can't. I even cannot locate the oxidizer.

8. Feb 16, 2005

### dextercioby

Sulphate ion is the oxydizer (the sulphur gets reduced).Okay,i'll right the reaction and u'll do the REDOX THING...

$$Cu+2H_{2}SO_{4}\rightarrow CuSO_{4}+SO_{2}\uparrow +2H_O$$

You can solve your problem now,but it would be nice if you figured out how this reaction is trully going...

Daniel.