Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{m_{Ag}} = 19,2 - 8,4 = 10,8g\\
\% Fe = \dfrac{{8,4}}{{19,2}} \times 100\% = 43,75\% \\
\% Ag = 100 - 43,75 = 56,25\% \\
b)\\
{n_{Ag}} = \dfrac{{10,8}}{{108}} = 0,1mol\\
2Ag + 2{H_2}S{O_4} \to A{g_2}S{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{{n_{Ag}}}}{2} = 0,05mol\\
{V_{S{O_2}}} = 0,05 \times 22,4 = 1,12l\\
{n_{A{g_2}S{O_4}}} = \dfrac{{{n_{Ag}}}}{2} = 0,05mol\\
{m_{A{g_2}S{O_4}}} = 0,05 \times 312 = 15,6g
\end{array}\)