Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
A + 2{H_2}O \to A{(OH)_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,2688}}{{22,4}} = 0,012mol\\
{n_A} = {n_{{H_2}}} = 0,012mol\\
{M_A} = \dfrac{m}{n} = \dfrac{{1,644}}{{0,012}} = 137dvC\\
\Rightarrow A:Bari(Ba)\\
b)\\
{n_{Ba{{(OH)}_2}}} = {n_{Ba}} = 0,012mol\\
{m_{Ba{{(OH)}_2}}} = 0,012 \times 171 = 2,052g\\
{m_{{\rm{dd}}spu}} = 1,644 + 50 - 0,012 \times 2 = 51,62g\\
C{\% _{Ba{{(OH)}_2}}} = \dfrac{{2,052}}{{51,62}} \times 100\% = 3,975\% \\
2)\\
a)\\
M + 2{H_2}O \to M{(OH)_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_M} = {n_{{H_2}}} = 0,1mol\\
{M_M} = \dfrac{4}{{0,1}} = 40dvC\\
\Rightarrow M:Canxi(Ca)\\
b)\\
{n_{Ca{{(OH)}_2}}} = {n_{Ca}} = 0,1mol\\
{C_{{M_{Ca{{(OH)}_2}}}}} = \dfrac{{0,1}}{{0,1}} = 1M
\end{array}\)