Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Zn}} = 64,36\% \\
\% {m_{Al}} = 35,64\% \\
b)\\
{C_\% }A{l_2}{(S{O_4})_3} = 44,79\% \\
{C_\% }ZnS{O_4} = 31,63\% \\
c)\\
{V_{S{O_2}}} = 2,24l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Zn + 2{H_2}S{O_4} \to ZnS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
hh:Zn(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
a + 1,5b = 0,45\\
65a + 27b = 15,15
\end{array} \right.\\
 \Rightarrow a = 0,15;b = 0,2\\
\% {m_{Zn}} = \dfrac{{0,15 \times 65}}{{15,15}} \times 100\%  = 64,36\% \\
\% {m_{Al}} = 100 - 64,36 = 35,64\% \\
b)\\
{n_{{H_2}S{O_4}}} = 2{n_{S{O_2}}} = 0,9\,mol\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{0,9 \times 98}}{{98\% }} = 90g\\
{m_{ddspu}} = 15,15 + 90 - 0,45 \times 64 = 76,35g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1\,mol\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,15\,mol\\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,1 \times 342}}{{76,35}} \times 100\%  = 44,79\% \\
{C_\% }ZnS{O_4} = \dfrac{{0,15 \times 161}}{{76,35}} \times 100\%  = 31,63\% \\
c)\\
{m_{Zn}} = 10,1 \times 64,36\%  = 6,5g\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1\,mol\\
{n_{S{O_2}}} = {n_{Zn}} = 0,1\,mol\\
{V_{S{O_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)