Consider the reaction:
Cu(s)+4HNO3(aq)→2NO2(g)+Cu(NO3)2(aq)+2H2O(g)
What volume of NO2 measured at 2.0 atm and 27∘C can be produced by the reaction of 0.500 mol copper with excess nitric acid according to the equation above?
Cu(s) + 4HNO3(aq)→2NO2(g) + Cu(NO3)2(aq)+2H2O(g)
When 1 mole of copper is used, two moles of NO2 and two moles of H2O are produced.
So, for 0.5 moles of copper, 1 mole of NO2 is produced.
PV=nRT
⇒(2)V=(1)(0.0821)(300)
⇒V=12 L