a,PTHH Cu +4HNO3--->Cu(NO3)2+2NO2+2H2O(Vì Fe ko pứ vs dung dịch HNO3 đặc,nguội)
Theo PTHH : nCu=1/2nNO2=1/2 x $\frac{4,48}{22,4}$= 0,1 mol
mCu=0,1x64=6,4g
%mCu=(6,4/20)x100=32%
mFe=20-6,4=13,6g
%mFe=100-32=68%
b,Theo PTHH:nHNO3=4nCu=4x0,1=0,4mol
=>VHNO3=0,4/8=0,05l