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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.

  • Chemistry -

    .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.

  • Chemistry -

    I think 0.5/63.54 = 0.00787 mols Cu.

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