Chemistry help again...

When 5.37g of the black copper(II)oxide is mixed with excess sulfuric acid, the solution turns a clear blue, indicating the formation of copper(II) sulfate. How many grams of copper(II) sulfate will be formed in this double displacement reaction???

ALSO: If an excess of zinc metal is placed in a copper (II) sulfate solution, the zinc will displace the copper. If 10.8g of copper(II) sulfate is reacted, how many grams of copper metal will be recovered from the solution? On this one i know the equation: Zn + CuSO4--> ZnS)4 + Cu.

PLEASE HELP MEEEE! :)

I'll do one and give you the answer for the other.
Problem One
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Step One
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Balance the equation. This is always the first step.

Cu(ii)O + H2SO4 ===> Cu(ii)SO4 + H2O

It is balanced.

Step Two
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Find the moles of Cu(ii)O
Cu = 63.5
O = 16
Molar Mass = 79.5 grams / mole

1 mole CuO = 79.5 grams
x mole CuO = 5.37 grams

x = 5.37 / 79.5 = 0.0675 moles of CuO

Step Three
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Find the moles of CuSO4 produced. This is what the balanced equation is for. If there is no number infront of a chemical, the number is assumed ot be one.

1 mole of CuO produces 1 mole of CuSO4
0.0675 moles of CuO produces x moles of CuSO4

x = 0.0675 moles. I know this is trivial, but it may not be in your next question.

Step Four
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Find the mass of CuSO4

1 mole CuSO4 is
Cu = 63.5
S = 32
O4 = 64 (from 4 * 16)
Molar Mass = 159.5 grams.

1 mole of CuSO4 = 159.5 grams CuSO4
0.0675 moles CuSO4 = x grams.

x = 159.5 * 0.0675
x = 10.77 grams.

Problem II
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Zn + CuSO4 ===> ZnSO4 + Cu

It is balaned so step two is again trivial.

The answer is 4.29 grams of Cu.