Đáp án:
\(\begin{array}{l}
12)\\
Al,Ga\\
13)\\
X:Magie(Mg),Y:B{\rm{ar}}i(Ba)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
12)\\
2R + 6HCl \to 2RC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol \Rightarrow {n_R} = \frac{{0,3 \times 2}}{3} = 0,2\,mol\\
{M_R} = \dfrac{{8,8}}{{0,2}} = 44(g/mol)\\
\Rightarrow hh:Al,Ga\\
13)\\
A + {H_2}S{O_4} \to AS{O_4} + {H_2}\\
{n_A} = {n_{{H_2}}} = \dfrac{{2,016}}{{22,4}} = 0,09mol\\
{M_A} = \dfrac{{2,64}}{{0,09}} = 29,33(g/mol)\\
{M_X} < {M_A} = 29,33 < {M_Y}\\
\Rightarrow X:Magie(Mg),Y:B{\rm{ar}}i(Ba)
\end{array}\)
