[bpeck8]

I am really lost on how to approach these problems.

For the following questions, assume that you are starting with 0.50g of copper.

problem 1. oxidation of Cu
               Concentrated nitric acid has a concentration of 15.9M. Calculate the amount of acid required to complete reaction 1.

Problem 2. formation of Cu(OH)2
           a. how much 3.0 M NaOH is needed to react completely with the copper in solution?
           b. How much grams of Cu(OH)2 should be produced?

problem 3. Dehydration of Cu(oh)2
               If Cu(OH)2 is heated to dryness, how many grams of pure copper oxide should be produced in reaction 3?

Problem 4. Reaction of CuO with sulfuric acid
               How many mL of 6.0 M H2SO4 are needed to react completely with the CuO?

Problem 5. Formation of Cu(s)
               How many grams of zinc are needed to yield 100% of the starting copper?

If you guys need additional information to help me out from me i can provide, just let me know.

Thanks,
Brent

[bpeck8]

I think i solved problem 2.

Cu + 2NaOH ---> Cu(OH)2 + 2Na

Part A) 3.0 M NaOH per L
           0.0078 mol Cu

           0.0078 mol Cu   X   2 mol NaOH  X   1 L             = 0.0052 L NaOH
                                      1 mol Cu         3 mol NaOH 

           0.0052 L NaOH  X  1 mL     = 5.2 mL NaOH
                                     0.001 L

Part B) 0.0078 mol Cu  X  1 mol Cu(OH)2   X   98 g             = 0.7644 g Cu(OH)2  this will be rounded off to .76 g Cu(OH)2
                                   1 mol Cu               1 mol Cu(OH)2

[bpeck8]

I think i might have done 1 also, not sure about the equation though.

Cu + 2HNO3 ----> Cu(NO3)2 + H2

15.9 M HNO3 Per L
0.0078 mol Cu

0.0078 mol Cu   X   2 mol HNO3   X       1 L             X   1 mL    = 0.98 mL HNO3
                           1 mol Cu         15.9 mol HNO3        0.001 L

[bpeck8]

I also think i might have done number 3 right

Cu(OH)2 ----> CuO + H20

0.0078 mol Cu   X   79 g CuO    =  .62 g CuO
                           1 mol CuO

[bpeck8]

Ok what about number 4

CuO + H2SO4 ----> CuSO4 + H2O

6.0 M H2SO4 per L
0.0078 mol Cu

0.0078 mol Cu  X  1 mol H2SO4   X   1 L               X   1 mL     =  1.3 mL H2SO4
                         1 mol Cu             6 mol H2SO4       0.001 L

[bpeck8]

Lastly number 5

Cu + Zn ----> Cu + Zn

0.0078 mol Cu   X   1 mol Zn   X   65 g Zn   = .51 g Zn
                           1 mol Cu       1 mol Zn


Sorry for so many posts,

Can Anyone confirm my solutions or tell me where i went wrong anywhere?

Thanks,
Brent

[Borek]

Not bad, a is off though, dissolution of copper nitric acid needs a redox reaction.