(1)n(Cu) == 1.92/64 ==0.03mol
n(HNO3)==0.2*0.25 ==0.05mol
n(H2SO4)==0.1*0.25==0.025mol
(2)3Cu + 2NO3- + 8H+ == 3Cu2+ + 2NO + 4H2O
---0.03--0.05---0.1----0.03mol
---0.03--0.02---0.08----0.03mol
(3)所得溶液中Cu*2+物质的量浓度==0.03/0.25==0.12mol/L