Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
b)\\
{n_{S{O_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45mol\\
hh:Al(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11\\
\dfrac{3}{2}a + \dfrac{3}{2}b = 0,45
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
\% Al = \dfrac{{5,4}}{{11}} \times 100\% = 49,09\% \\
\% Fe = 100 - 49,09 = 50,91\% \\
c)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = {n_{Fe}} + \frac{3}{2}{n_{Al}} = 0,4mol\\
{V_{{O_2}}} = 0,4 \times 22,4 = 8,96l\\
d)\\
{n_{NaOH}} = 0,5 \times 2 = 1mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{1}{{0,45}} = 2,22\\
T > 2 \Rightarrow\text{Phản ứng taon ra muối } N{a_2}S{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3}\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,45mol\\
{m_{N{a_2}S{O_3}}} = 0,45 \times 126 = 56,7g\\
e)\\
{n_{{H_2}S{O_4}}} = 3{n_{Al}} + 3{n_{Fe}} = 0,9mol\\
{m_{{H_2}S{O_4}}} = 0,9 \times 98 = 88,2g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{88,2 \times 100}}{{98}} = 90g
\end{array}\)
