Đáp án:
46,3 ml
Giải thích các bước giải:
\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O(1)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O(2)\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O(3)\\
n{H_2}S{O_4}(1,2) = 2nS{O_2} = \dfrac{{5,6}}{{22,4}} \times 2 = 0,5\,mol\\
nNaOH = 0,5 \times 1 = 0,5\,mol\\
\Rightarrow n{H_2}S{O_4}(3) = \dfrac{{0,5}}{2} = 0,25\,mol\\
n{H_2}S{O_4} = 0,5 + 0,25 = 0,75\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,75 \times 98}}{{98\% }} = 75g\\
V{H_2}S{O_4} = \dfrac{{75}}{{1,62}} = 46,3ml
\end{array}\)