PTHH: 2Al + 3$H_{2}$$SO_{4}$ -> $Al_{2}$($SO_{4}$)${_3}$ + 3$H_{2}$ (1)
Fe + $H_{2}$$SO_{4}$ -> $FeSO_{4}$ + $H_{2}$ (2)
Đặt $n_{Al}$ = a (mol) , $n_{Fe}$ = b (mol)
=> 27a + 56b = 11,1 (a)
Từ PTHH (1) =>$n_{H2(1)}$ = $\frac{3}{2}$ $n_{Al}$ = $\frac{3}{2}$a (mol)
Từ PTHH (2) => $n_{H2(2)}$ = $n_{Fe}$ = b (mol)
=> $n_{H2}$ = $\frac{3}{2}$a + b = $\frac{6,72}{22,4}$ (b)
Giải (a), (b) => $\left \{ {{a=0,1} \atop {b=0,15}} \right.$
=> %$m_{Al}$ $\frac{0,1*27}{11,1}$ *100% ≈ 24,32%
=> %$m_{Fe}$ = 100% - 24,32% = 75,68%
b. Từ PTHH (1) => $n_{H2SO4(1)}$ = $\frac{3}{2}$ $n_{Al}$ =0,15 (mol)
Từ PTHH (2) => $n_{H2SO4(2)}$ = $n_{Fe}$ =0,15 (mol)
=> $n_{H2SO4}$ = 0,3 (mol)
=> $C_{M H2SO4}$ = $\frac{0,3}{300:100}$ = 1(M)