Đáp án:
$a/$
Fe + H2SO4 --> FeSO4 + H2
x x x x
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
y 3/2y 1/2y 3/2y
$nH2 = 4,48/22,4 = 0,2 (mol)$
Ta có:
$\left \{ {{56x + 27y=5,5} \atop {y+3/2y=0,2}} \right.$ => $\left \{ {{x=0,05} \atop {y=0,1}} \right.$
$b/$
%$mFe = $$\frac{0,05.56}{5,5}$$.100 = 50,91$%
%$mAl = 100 - 50,91 = 49,09$%