1 mole Cu reduces ------ moles of dil.HNO3

3Cu+8HNO3(dil.)→3Cu(NO3)2+2NO+4H2O    All the 3 moles of Cu Oxidises to Cu+2   Only 2 moles HNO3(of the total 8 moles) reduces to NO    âˆ´3 moles of Cu reduces 2 moles of  dil.HNO3    1 mole of Cu reduces 'X' moles of dil.HNO3                      X = 2/3Alternate method :                            Cu0+N+5O3- â†’Cu+2+N+2OTotal increase in ox. state of Cu = 2Total decrease in ox. state of 'N' = 3∴3 moles of Cu reduces 2 moles of  dil.HNO3     1 mole of Cu reduces 'X' moles of dil.HNO3                    X = 2/3