Asked by Morgan
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.
Answers
Answered by
Steve
.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.
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.
Answered by
DrBob222
I think 0.5/63.54 = 0.00787 mols Cu.
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