Giải thích các bước giải:

Ag không tác dụng với H2SO4 loãng

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
a)\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
 \to {m_{Al}} = 0,2 \times 27 = 5,4g\\
 \to {m_{Ag}} = 16,2 - 5,4 = 10,8g \to {n_{Ag}} = 0,1mol\\
b)\\
2Ag + 2{H_2}S{O_4} \to A{g_2}S{O_4} + 2{H_2}O + S{O_2}\\
{n_{S{O_2}}} = \dfrac{1}{2}{n_{Ag}} = 0,05mol\\
 \to {V_{S{O_2}}} = 0,05 \times 22,4 = 1,12l\\
c)\\
{n_{NaOH}} = 0,1mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,1}}{{0,05}} = 2
\end{array}\)

-> Tạo 1 muối: \(N{a_2}S{O_3}\)

\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,05mol\\
 \to {m_{N{a_2}S{O_3}}} = 0,05 \times 126 = 6,3g\\
d)\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 6{H_2}O + 3S{O_2}\\
2Ag + 2{H_2}S{O_4} \to A{g_2}S{O_4} + 2{H_2}O + S{O_2}\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Al}} + \dfrac{1}{2}{n_{Ag}} = \dfrac{3}{2} \times 0,2 + \dfrac{1}{2} \times 0,1 = 0,35mol\\
 \to {V_{S{O_2}}} = 0,35 \times 22,4 = 7,84l
\end{array}\)