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Copper metal reacts with concentrated nitric acid according to the following balanced equation:

3Cu(s) + 8HNO3(aq) ¨ 3Cu(NO3)2(s) + 2NO(g) + 4H2O(l)

Calculate the mass in grams of the excess reagent remaining after the complete reaction of 4.63 g of Cu with 9.26 g of HNO3. Round your answer to 2 decimal places.

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    1. You have the balanced equation.
    2. Convert g Cu to mols Cu.
    3. Convert g HNO3 to mols HNO3.
    4a. Convert mols Cu to mols Cu(NO3)2.
    4b. Convert mols HNO3 to mols Cu(NO3)2.
    4c. Choose the smaller mols of Cu(NO2)2 2. The reagent producing that number (either 4a or 4b) will be the limiting reagent.
    5. Convert mols of the limiting reagent to mols of the other reagent. That will be how much of the other reagent (the excess reagent) is used.
    6. Convert mols of the excess reagent used to grams, subtract from the initial mass present, and that will give you the amount remaining after reaction.
    Post your work if you need more help.

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