$a)PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$2Ag+2H_2SO_4\to Ag_2SO_4+SO_2\uparrow+2H_2O$

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol);n_{SO_2}=\dfrac{1,12}{22,4}=0,05(mol)$

Từ 2 PT $\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol);n_{Ag}=2n_{SO_2}=0,1(mol)$

$\Rightarrow m_{Al}=0,2.27=5,4(g);m_{Ag}=0,1.108=10,8(g)$

$\Rightarrow \%m_{Al}=\dfrac{5,4}{5,4+10,8}.100\%=33,33\%$

$\Rightarrow \%m_{Ag}=100-33,33=66,67\%$

$b)m_{SO_2}=0,05.32=1,6(g);m_{H_2}=0,3.2=0,6(g)$

$\Rightarrow \overline{M_{hh}}=\dfrac{1,6+0,6}{0,05+0,3}=\dfrac{44}{7}(g/mol)$

$\Rightarrow d_{M_{hh}/O_2}=\dfrac{\dfrac{44}{7}}{32}\approx 0,196$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
2Ag + 2{H_2}S{O_4} \to A{g_2}S{O_4} + S{O_2} + 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Al}} = 0,3 \times \dfrac{2}{3} = 0,2\,mol\\
{n_{S{O_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
{n_{Ag}} = 2{n_{S{O_2}}} = 0,1\,mol\\
\% {m_{Al}} = \dfrac{{0,2 \times 27}}{{0,2 \times 27 + 0,1 \times 108}} \times 100\%  = 33,33\% \\
\% {m_{Ag}} = 100 - 33,33 = 66,67\% \\
b)\\
{M_{hh}} = \dfrac{{0,3 \times 2 + 0,05 \times 64}}{{0,3 + 0,05}} = \dfrac{{76}}{7}\,g/mol\\
{d_{hh/{O_2}}} = \dfrac{{\dfrac{{76}}{7}}}{{32}} = 0,339
\end{array}\)