Đáp án:
b) 20,59% 79,41%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
P1:\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
P2:\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Ag + 2{H_2}S{O_4} \to A{g_2}S{O_4} + S{O_2} + 2{H_2}O\\
b)\\
n{H_2} = \dfrac{{2,688}}{{22,4}} = 0,12\,mol \Rightarrow nFe = n{H_2} = 0,12\,mol\\
nS{O_2} = \dfrac{{7,392}}{{22,4}} = 0,33\,mol\\
\Rightarrow nAg = 2 \times (0,33 - 0,12 \times 1,5) = 0,3\,mol\\
\% mFe = \dfrac{{0,12 \times 56}}{{0,12 \times 56 + 0,3 \times 108}} \times 100\% = 20,59\% \\
\% mAg = 100 - 20,59 = 79,41\%
\end{array}\)