Đáp án:
\(\begin{array}{l}
a)\\
hh:Natri(Na),Kali(K)\\
b)\\
\% {m_{Na}} = 37,1\% \\
\% {m_K} = 62,9\% \\
c)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 32,67g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2R + 2{H_2}O \to 2ROH + {H_2}\\
{n_R} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
{n_R} = 2{n_{{H_2}}} = 0,1\,mol\\
{M_R} = \dfrac{{3,1}}{{0,1}} = 31g/mol\\
\Rightarrow hh:Natri(Na),Kali(K)\\
b)\\
hh:Na(a\,mol),K(b\,mol)\\
\left\{ \begin{array}{l}
23a + 39b = 3,1\\
a + b = 0,1
\end{array} \right.\\
\Rightarrow a = b = 0,05\\
\% {m_{Na}} = \dfrac{{0,05 \times 23}}{{3,1}} \times 100\% = 37,1\% \\
\% {m_K} = 100 - 37,1 = 62,9\% \\
c)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,05}}{2} + \dfrac{{0,05}}{2} = 0,05\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,05 \times 98}}{{15\% }} = 32,67g
\end{array}\)
