Đáp án:
\(\begin{array}{l}
a)\\
hh:Natri(Na),Kali(K)\\
b)\\
\% {m_{Na}} = 56,47\% \\
\% {m_K} = 43,53\% \\
c)\\
{C_M}NaOH = 2,75M\\
{C_M}KOH = 1,25M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2R + 2{H_2}O \to 2ROH + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_R} = 2{n_{{H_2}}} = 0,8\,mol\\
{M_R} = \dfrac{{22,4}}{{0,8}} = 28(g/mol)\\
\Rightarrow hh:Natri(Na),Kali(K)\\
b)\\
hh:Na(a\,mol),K(b\,mol)\\
23a + 39b = 22,4\\
a + b = 0,8\\
\Rightarrow a = 0,55;b = 0,25\\
\% {m_{Na}} = \dfrac{{0,55 \times 23}}{{22,4}} \times 100\% = 56,47\% \\
\% {m_K} = 100 - 56,47 = 43,53\% \\
c)\\
{n_{NaOH}} = {n_{Na}} = 0,55\,mol\\
{n_{KOH}} = {n_K} = 0,25\,mol\\
{C_M}NaOH = \dfrac{{0,55}}{{0,2}} = 2,75M\\
{C_M}KOH = \dfrac{{0,25}}{{0,2}} = 1,25M
\end{array}\)
