Đáp án:

$a,\ NaHCO_3;\ Na_2CO_3\\ b,\\ \%m_{NaHCO_3}=61,31\%\\ \%m_{Na_2CO_3}=38,69\%$

Giải thích các bước giải:

$a,\\n_{CO_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ n_{NaOH}=0,05.2=0,1(mol)\\ n_{HCl}=0,5.1=0,5(mol)\\ Đặt\ \begin{cases}n_{MHCO_3}=x(mol)\\n_{M_2CO_3}=y(mol)\end{cases}\\ MHCO_3+HCl\to MCl+H_2O+CO_2\ (1)\\ M_2CO_3+2HCl\to MCl_2+H_2O+CO_2\ (2)\\ n_{CO_2}=n_{MHCO_3}+n_{M_2CO_3}=x+y(mol)\\ \Rightarrow x+y=0,3\ (1)\\ NaOH+HCl\to NaCl+H_2O\\ n_{HCl\ dư}=n_{NaOH}=0,1(mol)\\ \Rightarrow n_{HCl\ (1),(2)}=0,5-0,1=0,4(mol)\\ n_{HCl\ (1),(2)}=n_{MHCO_3}+2.n_{M_2CO_3}=x+2y(mol)\\ \Rightarrow x+2y=0,4\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,2\\y=0,1\end{cases}\\ \Rightarrow 0,2.(M+61)+0,1.(2M+60)=27,4\\ \Rightarrow M=23\\ \to M:\ Na\\ \to \text{2 muối gồm}:\ NaHCO_3;\ Na_2CO_3$

$b,\\ \%m_{NaHCO_3}=\dfrac{0,2.84}{27,4}.100\%=61,31\%\\ \%m_{Na_2CO_3}=\dfrac{0,1.106}{27,4}.100\%=38,69\%$