Đáp án:
\(\begin{array}{l}
M:B{\rm{er}}i(Be)\\
\% {m_{Be}} = 65,22\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
M + {H_2}S{O_4} \to MS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{26,88}}{{22,4}} = 1,2\,mol \Rightarrow {n_{hh}} = {n_{{H_2}}} = 1,2\,mol\\
{M_{hh}} = \dfrac{{13,8}}{{1,2}} = 11,5\,g/mol \Rightarrow {M_M} < 11,5 \Rightarrow M:B{\rm{er}}i(Be)\\
hh:Be(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
9a + 24b = 13,8\\
a + b = 1,2
\end{array} \right.\\
\Rightarrow a = 1;b = 0,2\\
\% {m_{Be}} = \dfrac{{1 \times 9}}{{13,8}} \times 100\% = 65,22\% \\
\% {m_{Mg}} = 100 - 65,22 = 34,78\%
\end{array}\)
