bai 10
a , Al+6HN03=Al(N03)3+3N02+2H20
Cu+4HNO3===>Cu(NO3)2+2NO2+2H20
b, nNO2=17,92/22,4=0,8 mol
Al(0)===>Al(3+)+3e
x.........................3x
Cu(0)===>Cu(2+)+2e
y............................2y
N(+5)+1e===>N(+4)
0,8.........0,8
e cho =e nhận
=>3x+2y=0,8
lại có 27x+64y=11,8
giaỉ hpt {x=0,2
             {y=0,1
=>mAl=27.0,2=5,4g
=>%Al=5,4/11,8.100%=45,76%
mCu=64.0,1=6,4g
%Cu=6,4/11,8.100%=54,24%