AP CHEMISTRY HELP! An 8.00 g sample of a copper ore was dissolved in a acid solution and then reacted with?

Step One
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Find the moles of NO3- on the left side of the equation.
v = 71.5 mL = 0.0715 L
C = 0.155 M

n = C*V = 0.0111 moles

Step Two
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Find the number of moles of copper

3 moles of copper requires 8 moles of N03-
x moles of copper requires 0.0111 moles of N03-

8x = 3*0.0111
8x = 0.0333
x = 0.00416 moles of copper

Step Three
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Find the grams of copper.
n = 0.00416 mol
given mass = x
Molar Mass = 63.5 grams / mol
x = n*MM
x = 0.00416 mol * 63.5 = 0.2639 grams

Step Four
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Find the % in the sample
% = (given mass/Sample Mass)*100
% = (0.2639/8.00) * 100 = 3.30%

First calculate the moles of NO3- (KNO3) from volume and molarity of KNO3.
Moles KNO3 = molarity x L of KNO3
= (0.155 M) (0.0715 L)
= (0.155 mole KNO3/L) (0.0715 L) = 0.0111 mole KNO3
Which means the solution contains 0.0111 mole K+ and 0.0111 mole NO3- ions.
Use the stoichiometry from the equation to calculate the moles Cu and grams Cu using the following setup.

0.0111 mole NO3- (3 mole Cu/8 mole NO3-) (63.55 g Cu/1 mole Cu) = 0.265 g Cu

This means that 8.00g of ore contains 0.265 g Cu.
Precent Cu in the ore = (0.265 g /8.00 g) x 100% = 3.31%