posted by Morgan on .
You have .5g of copper. If the nitric acid is 16 M HNO3, how many milliliters of nitric acid will react exactly with that amount of copper. I know that when I get that volume I will triple it so that there will be excess acid and the copper will be the limiting reactant.
.5g Cu is .000787 moles Cu
16M HNO3 is 16 moles/L
You need 2 moles of HNO3 for each mole of Cu, so that's .00157 moles HNO3
.00157 mole / 16mole/L = .0000981 L = .0981mL
Seems like an awfully small volume. Better check my math.
I think 0.5/63.54 = 0.00787 mols Cu.