Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
hh:Al(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11\\
\dfrac{3}{2}a + b = 0,4
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
b)\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
TH1:\text{Thu được kết tủa lớn nhất}\\
FeS{O_4} + 2NaOH \to Fe{(OH)_2} + N{a_2}S{O_4}\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
{n_{NaOH}} = 2{n_{FeS{O_4}}} + 6{n_{A{l_2}{{(S{O_4})}_3}}} = 0,8mol\\
{V_{NaOH}} = \dfrac{{0,8}}{2} = 0,4l = 400ml\\
TH2:\text{Thu được kết tủa bé nhất}\\
FeS{O_4} + 2NaOH \to Fe{(OH)_2} + N{a_2}S{O_4}\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
{n_{Al{{(OH)}_3}}} = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,2mol\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
{n_{NaOH}} = 2{n_{FeS{O_4}}} + 6{n_{A{l_2}{{(S{O_4})}_3}}} + {n_{Al{{(OH)}_3}}} = 1mol\\
{V_{NaOH}} = \dfrac{1}{2} = 0,5l = 500ml
\end{array}\)
