Question

3. Cu(s) + 4HNO3(aq) --> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

a)Write the net ionic equation for the reaction of concentrated nitric acid with solid copper

b) calculate the volume of concentrated nitric acid (16M) required to react with 0.37 g of Cu(s)

c) If you use 1.69 mL (an excess) of HNO3, what volume of 5.0M NaOH will you need to neutralize the solution after the copper is compeltely dissolved?

Answer 1

3)

a)

molecular equation : Cu(s) + 4HNO3(aq) -----------> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Ionic Equation: Cu(s) + 4 H+ (aq) + 4 NO₃^-(aq) ——> Cu2+ (aq) + 2 NO₃^- (aq) + 2 NO(g) + 2H₂O(l)

Net ionic equation: Cu(s) + 4H+ (aq) + 4 NO₃^-(aq) ——> Cu2+ (aq) + 2 NO₃^- (aq)

b)

concentration = 16 M

moles of Cu = 0.37 / 63.55 = 5.82 x 10^-3

1 mol Cu   ------------> 4 mol HNO3

5.82 x 10^-3 mol Cu   ----------> ??

moles of HNO3 = 5.82 x 10^-3 x 4 / 1 = 0.0233

volume of HNO3 = 0.0233 x 16 = 0.373 L

volume of HNO3 = 372.6 mL

c)

moles of HNO3 = 1.69 x 16 / 1000 = 0.027 mol

HNO3   +   NaOH   --------------> NaNO3 + H2O

moles of HNO3 = moles of NaOH

moles of NaOH = 0.027 mol

volume of NaOH = 0.027 x 5 = 0.135 L

volume of NaoH = 135 mL