Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Na}} = 54,12\% \\
\% {m_K} = 45,88\% \\
b)\\
{V_{HCl}} = 0,15l\\
m = 19,15g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
a)\\
2X + 2HCl \to 2XCl + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow {n_X} = 0,15 \times 2 = 0,3\,mol\\
{M_X} = \dfrac{{8,5}}{{0,3}} = 28,33(g/mol)\\
\Rightarrow hhX:Na(a\,mol),K(b\,mol)\\
23a + 39b = 8,5\\
a + b = 0,3\\
\Rightarrow a = 0,2;b = 0,1\\
\% {m_{Na}} = \dfrac{{0,2 \times 23}}{{8,5}} \times 100\% = 54,12\% \\
\% {m_K} = 100 - 54,12 = 45,88\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3mol\\
{V_{HCl}} = \dfrac{{0,3}}{2} = 0,15l\\
m = {m_{NaCl}} + {m_{KCl}} = 0,2 \times 58,5 + 0,1 \times 74,5 = 19,15g
\end{array}\)
