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:

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

a,

$n_{CO_2}=\dfrac{2,8}{22,4}=0,125(mol)$

$M_2CO_3+2HCl\to 2MCl+CO_2+H_2O$

$MHCO_3+HCl\to MCl+CO_2+H_2O$

$\to n_{hh}=n_{CO_2}=0,125(mol)$

$\to \overline{M}_{hh}=\dfrac{14,4}{0,125}=115,2$

Ta có:

$M_M>1$

$\to M_M+M_M>M_M+1$

$\to 2M_M>M_M+1$        

$\to 2M_M+60>M_M+1+60$

$\to M_{M_2CO_3}>M_{MHCO_3}$

$\to M_M+61<115,2<2M_M+60$

$\to 27,6<M_M<54,2$

$\to M_M=39(K)$

Vậy KL là kali 

b,

Gọi $x$, $y$ là số mol $K_2CO_3$, $KHCO_3$

$\to 138x+100y=14,4$

Mặt khác: $x+y=0,125$

$\to x=0,05; y=0,075$

$\%m_{K_2CO_3}=\dfrac{0,05.138.100}{14,4}=47,9\%$

$\%m_{KHCO_3}=52,1\%$