Stoichiometry, Limiting Reagents, Solutions?

Question

Copper metal reacts with nitric acid to produce copper nitrate, water and nitric oxide (in the reaction shown below). An inorganic chemist placed 100.0 mg of copper powder into a reaction flask with 100.0 mL of 0.037 M nitric acid (HNO3). After allowing the reaction to proceed to completion, he evaporated the contents of the flask to dryness. To the correct number of significant figures, how much solid material remains in the flask?

HNO3(aq) + Cu(s) Cu(NO3)2(s) + H2O(l) + NO(g

Answer 1

Find the moles of HNO3 (0.1 L * 0.037 moles/l)= "A" moles.
Since 1 mole of HNO3 reacts with 1 g-atom Cu, that much Cu is removed.
Multiply "A" by the atomic weight of Cu to get the Cu removed in mg. Call that "B"
The amount remaining is then 100 mg - B.

Answer 2

Alright, after scribbling down my thoughts on some paper, I came up with 299.4 mg of solid Copper Nitrate.

I did this by saying that we reacted 3.7 mmol of nitric acid with the limiting reagant, Cu (1.6 mmol) and that yeilded 1.6 mmol of copper nitrate... at 186.9 mg/mol.


No one else had an answer up when I started typing this so I wonder how I did....