Balance the following reaction, then calculate the number of moles of water (l = liquid) produced when 50g of Cu (s=solid) are supplied with 3.0 moles of HNOthree (aq=solution) Is copper or nitric acid the limiting reagent?

Cu(s) + HNOthree (aq) ----> Cu(NOthree)two (s) + NO(g) + H(two)O (l)

1. 50 grams of Cu. The atomic weight of Cu is 63.546 grams / mole. From these two pieces of information, you calculate the number of moles of copper you have by dividing:

\Large \frac {grams}{(\frac {grams}{mole})} = mole

2. 3.0 moles of HNO3. They already gave you the number of moles of nitric acid. You don't have to calculate that.

3. Balance the equation (I've corrected it here). If your equation was given to you, then balance that one and use the coefficients in that equation to make your final decision. This is the reaction that occurs when you mix Copper with Nitric acid:

Cu(s) + 2HNO_3 \rightleftharpoons Cu(NO_3)_2 (aq) + H_2(g)

From the equation you'll note that you need 2 moles of HNO3 to react with 1 mole of Cu.

1. If you have less than 2 moles of HNO3 for every mole of Cu, then HNO3 is the limiting reagent.
2. If you have less than 1 mole of copper for every 2 moles of HNO3, then Cu is the limiting reagent.