another chem question i need help with?

Question

Nitrogen dioxide has opposing roles in our world. It is essential in the industrial production of nitric acid, but is also a contributor to acid rain and photochemical smog. Calculate the volume of nitrogen dioxide produced at 717.3 torr and 22.4°C by the reaction of 8.65 cm3 copper (d = 8.96 g/cm3) with 219.6 mL nitric acid (d = 1.42 g/cm3, 68.0% HNO3 by mass):

Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Answer 1

Step 1. Get your reactants in moles
For Copper: 8.65 cc*8.96 g/cc divided by mole wt=MC
For Nitric acid:
219.6 mL * 1.42 g/cc*0.68=MN

One of these reactants MAY be limiting. From the reaction the ratio of Cu/HNO3 = 1/4. If the actual ratio is lower, Cu will be the limiting reactant. If the actual ratio is higher, HNO3 will be the limiting reagent.

Based on the limiting reagent, compute the moles of NO2 produced. Use the reaction.

Now that you have moles of NO2, you can use the ideal gas law to compute volume.