\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right);n_{NaOH}=0,05.2=0,1\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{R_2CO_3}=x\left(mol\right)\\n_{RHCO_3}=y\left(mol\right)\end{matrix}\right.\)
PTHH:
\(R_2CO_3+2HCl\rightarrow2RCl+CO_2\uparrow+H_2O\)
x-------->2x-------------------->x
\(RHCO_3+HCl\rightarrow RCl+CO_2\uparrow+H_2O\)
y------------>y---------------->y
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,1------>0,1
=> \(\left\{{}\begin{matrix}x+y=0,3\\x+y+0,1=0,5\end{matrix}\right.\) => x = 0,1; y = 0,2
=> \(0,1.\left(2M_R+60\right)+0,2.\left(M_R+61\right)=27,4\)
=> MR = 23 (g/mol)
=> R là Na
=> 2 muối là Na2CO3, NaHCO3
\(\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,1.106}{27,4}.100\%=38,7\%\\\%m_{NaHCO_3}=100\%-38,7\%=61,3\%\end{matrix}\right.\)