Đáp án:

\(\begin{array}{l}
a)\\
{m_{Fe}} = 5,6g\\
{m_{Cu}} = 6,4g\\
b)\\
{V_{S{O_2}}} = 5,6l\\
c)\\
m = 29,3g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
{m_{Cu}} = 12 - 5,6 = 6,4g\\
b)\\
{n_{Cu}} = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = 0,1 \times \dfrac{3}{2} + 0,1 = 0,25\,mol\\
{V_{S{O_2}}} = 0,25 \times 22,4 = 5,6l\\
c)\\
{n_{NaOH}} = 0,4 \times 1 = 0,4\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,4}}{{0,25}} = 1,6\\
1 < T < 2 \Rightarrow \text{ Tạo cả 2 muối } \\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O(1)\\
N{a_2}S{O_3} + S{O_2} + {H_2}O \to 2NaHS{O_3}(2)\\
{n_{N{a_2}S{O_3}(1)}} = {n_{S{O_2}(1)}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{n_{N{a_2}S{O_3}(2)}} = {n_{S{O_2}(2)}} = 0,25 - 0,2 = 0,05\,mol\\
{n_{N{a_2}S{O_3}}} = 0,2 - 0,05 = 0,15\,mol\\
{n_{NaHS{O_3}}} = 2{n_{S{O_2}(2)}} = 0,1\,mol\\
m = 0,15 \times 126 + 0,1 \times 104 = 29,3g
\end{array}\)