Đáp án:
\(\begin{array}{l}
a)\\
Kali\\
b)\\
\% {m_{{K_2}C{O_3}}} = 47,92\% \\
\% {m_{KHC{O_3}}} = 52,08\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{M_2}C{O_3} + 2HCl \to 2MCl + C{O_2} + {H_2}O\\
MHC{O_3} + HCl \to MCL + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{2,8}}{{22,4}} = 0,125\,mol\\
{M_{hh}} = \dfrac{{14,4}}{{0,125}} = 115,2\,(g/mol)\\
\Rightarrow M + 61 < 115,2 < 2M + 60 \Rightarrow 27,5 < M < 54,2\\
\Rightarrow M = 39\,g/mol \Rightarrow M:Kali(K)\\
b)\\
hh:{K_2}C{O_3}(a\,mol),KHC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,125\\
138a + 100b = 14,4
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,075\\
\% {m_{{K_2}C{O_3}}} = \dfrac{{0,05 \times 138}}{{14,4}} \times 100\% = 47,92\% \\
\% {m_{KHC{O_3}}} = 100 - 47,92 = 52,08\%
\end{array}\)