Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Cu}} = 63,16\% \\
\% {m_{Fe}} = 36,84\% \\
b)\\
{m_{{\rm{dd}}HCl}} = 36,5g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1\,mol\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O(1)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O(2)\\
{n_{S{O_2}(1)}} = 0,1 \times \dfrac{3}{2} = 0,15\,mol\\
{n_{S{O_2}(2)}} = 0,3 - 0,15 = 0,15\,mol\\
{n_{Cu}} = {n_{S{O_2}(2)}} = 0,15\,mol\\
\% {m_{Cu}} = \dfrac{{0,15 \times 64}}{{0,15 \times 64 + 0,1 \times 56}} \times 100\% = 63,16\% \\
\% {m_{Fe}} = 100 - 63,16 = 36,84\% \\
b)\\
{n_{HCl}} = 2{n_{Fe}} = 0,2\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,2 \times 36,5}}{{20\% }} = 36,5g
\end{array}\)