Đáp án:

\(\begin{array}{l}
a)\\
m = 27,2g\\
b)\\
\% {m_{Na}} = 54,12\% \\
\% {m_K} = 45,88\% \\
c)\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = 0,3l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2R + 2{H_2}O \to 2ROH + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{{H_2}O}} = 2{n_{{H_2}}} = 0,6\,mol\\
m = 17 + 0,6 \times 18 - 0,3 \times 2 = 27,2g\\
b)\\
{n_R} = 2{n_{{H_2}}} = 0,6\,mol\\
{M_R} = \dfrac{{17}}{{0,6}} = 28,33\,g/mol \Rightarrow hh:Na,K\\
hh:Na(a\,mol),K(b\,mol)\\
\left\{ \begin{array}{l}
23a + 39b = 17\\
a + b = 0,6
\end{array} \right.\\
 \Rightarrow a = 0,4;b = 0,2\\
\% {m_{Na}} = \dfrac{{0,4 \times 23}}{{17}} \times 100\%  = 54,12\% \\
\% {m_K} = 100 - 54,12 = 45,88\% \\
c)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,4}}{2} + \dfrac{{0,2}}{2} = 0,3\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,3}}{1} = 0,3l
\end{array}\)