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2. In reaction (i), suppose you add 4.0 mL of 6 M nitric acid to a sphere of copper metal that weighs 0.65 grams. Which react

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Answer #1

You have the reaction:

2 HNO3 + Cu = Cu (NO3) 2 + H2

The moles of the reagents are calculated:

n HNO3 = M * V = 6 M * 0.004 L = 0.024 mol

n Cu = g / MM = 0.65 g / 63.5 g / mol = 0.010 mol

The limit reagent is Cu.

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