\(\begin{array}{l}
a)\\
Hh:\,Al:x(mol);Fe:y(mol)\\
\to 27x+56y=11(1)\\
2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\
Fe+H_2SO_4\to FeSO_4+H_2\\
Theo\,PT:\,1,5x+y=\frac{8,96}{22,4}=0,4(2)\\
(1)(2)\to x=0,2;y=0,1\\
\%m_{Al}=\frac{0,2.27}{11}.100\%\approx 49,09\%\\
\%m_{Fe}=100-49,09=50,91\%\\
b)\\
500ml=0,5l\\
Theo\,PT:\,n_{H_2SO_4}=n_{H_2}=0,4(mol)\\
\to C_{M\,H_2SO_4}=\frac{0,4}{0,5}=0,8M\\
c)\\
Theo\,PT:\,n_{Al_2(SO_4)_3}=0,1(mol);n_{FeSO_4}=0,1(mol)\\
\to C_{M\,Al_2(SO_4)_3}=C_{M\,FeSO_4}=\frac{0,1}{0,5}=0,2M
\end{array}\)

Đáp án+ Giải thích các bước giải:

            $\rm 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
            $\rm Fe+H_2SO_4\to FeSO_4+H_2$
$\rm a)$
$\rm Gọi\ n_{Al}:a(mol)\ ; n_{Fe}:b(mol)$
`=>`$\rm 27a+56b=11\ (1)$
$\rm n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$
`=>`$\rm 1,5a+b=0,4\ (2)$
`(1)(2)=>`$\rm a=0,2(mol)\ ; b=0,1(mol)$
`%`$\rm m_{Al}$`={0,2.27}/{11}.100%~~49,09%`
`%`$\rm m_{Fe}$`=100%-49,09%=50,91%`
$\rm b)$
$\rm Sửa : 500(ml)$`->`$\rm 500(g)$
$\rm n_{H_2SO_4}=n_{H_2}=0,4(mol)$
$\rm C$`%`$\rm _{dd\ H_2SO_4}$`=\frac{0,4.98}{500}.100%=7,84%`
$\rm c)$
$\rm m_{dd\ spu}=11+500-(0,4.2)=510,2(g)$
$\rm n_{Al_2(SO_4)_3}=0,5n_{Al}=0,1(mol)$
$\rm C$`%`$\rm _{dd\ Al_2(SO_4)_3}$`=\frac{0,1.342}{510,2}.100%=6,7%`
$\rm n_{FeSO_4}=n_{Fe}=0,1(mol)$
$\rm C$`%`$\rm _{dd\ FeSO_4}$`=\frac{0,1.152}{510,2}.100%=2,98%`