$ 2Al + 3H2SO4 loãng ---> Al2(SO4)3 + 3H2 $
x 3/2x x/2 3/2x mol
$ CuO + H2OS4 loãng ---> CuSO4 + H2O $
y y y y mol
n H2SO4 = 0,9.0,5= 0,45mol
$\left \{ {{27x+ 80y =17,4} \atop {3/2x+y= 0,45}} \right.$
$\left \{ {{x=0,2} \atop {y=0,15}} \right.$
m Al = 0,2.27= 5,4g
m CuO = 17,4- 5,4= 12g
%$m_{CuO/hh}$ $\frac{12.100%}{17,4}$ = $ 68,97% $
% m Al /hh = $100% - 68,97% $ = 31,03%
V H2= 3/2. 0,2.22,4= 6,72l
m muối = 0,1. 294 + 0,15. 160= 53,4g