Đáp án:
\(\begin{array}{l}
\% {m_{Mg}} = 37,5\% \\
\% {m_{Ca}} = 62,5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
X + 2HCl \to XC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_X} = {n_{{H_2}}} = 0,2\,mol\\
{M_X} = \dfrac{{6,4}}{{0,2}} = 32(g/mol)\\
\Rightarrow hh\,X:Magie(Mg);Can\,xi(Ca)\\
hh:Mg(a\,mol),Ca(b\,mol)\\
a + b = 0,2\\
24a + 40b = 6,4\\
\Rightarrow a = b = 0,1\,mol\\
\% {m_{Mg}} = \dfrac{{0,1 \times 24}}{{6,4}} \times 100\% = 37,5\% \\
\% {m_{Ca}} = 100 - 37,5 = 62,5\%
\end{array}\)