Đáp án:
\(\begin{array}{l}
M:Bari(Ba)\\
C{\% _{Ba{{(OH)}_2}}} = 26,9\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
R + 2{H_2}O \to R{(OH)_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_M} = {n_{{H_2}}} = 0,1mol\\
{M_M} = \dfrac{{13,7}}{{0,1}} = 137dvC\\
\Rightarrow M:Bari(Ba)\\
{m_{{\rm{dd}}spu}} = 13,7 + 50 - 0,1 \times 2 = 63,5g\\
{n_{Ba{{(OH)}_2}}} = {n_{{H_2}}} = 0,1mol\\
{m_{Ba{{(OH)}_2}}} = 0,1 \times 171 = 17,1g\\
C{\% _{Ba{{(OH)}_2}}} = \dfrac{{17,1}}{{63,5}} \times 100\% = 26,9\%
\end{array}\)