Đáp án:

 b) 20ml

Giải thích các bước giải:

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
hh:Fe(a\,mol),Cu(b\,mol)\\
n{H_2} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
nS{O_2} = \dfrac{{4,1216}}{{22,4}} = 0,184\,mol\\
nFe = n{H_2} = 0,05\,mol\\
 =  > nCu = 0,184 - \dfrac{{0,05 \times 3}}{2} = 0,109\,mol\\
m = mCu + mFe = 0,05 \times 2 \times 56 + 0,109 \times 2 \times 64 = 19,552g\\
\% mCu = 71,36\% ;\% mFe = 28,64\% \\
b)\\
n{H_2}S{O_4} = 0,05 \times 3 + 0,109 \times 2 = 0,368\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,368 \times 98}}{{98\% }} = 36,8g\\
V = \dfrac{{36,8}}{{1,84}} = 20ml
\end{array}\)