$2/\\ n_{NaOH}=0,8.1=0,8(mol)\\ \text{Coi muối chỉ có:}\ NaHCO_3\\ \xrightarrow{\text{BTNT Na}}\ n_{NaHCO_3}=n_{NaOH}=0,8(mol)\\ \to m_{NaHCO_3}=0,8.84=67,2(g)\\ \text{Coi muối chỉ có:}\ Na_2CO_3\\ \xrightarrow{\text{BTNT Na}}\ n_{Na_2CO_3}=\dfrac{1}{2}.n_{Na}=0,4(mol)\\ \to m_{Na_2CO_3}=0,4.106=42,4(g)\\ m_{Muối}<42,4<67,2\\ \to \text{NaOH dư, tạo muối trung hòa}\\ 2NaOH+CO_2\to Na_2CO_3+H_2O\\ Đặt\ n_{NaOH\ \text{phản ứng}}=x(mol)\\ \to n_{NaOH\ dư}=0,8-x(mol)\\ n_{Na_2CO_3}=\dfrac{1}{2}.n_{NaOH\ \text{phản ứng}}=0,5x(mol)\\ \to 0,5x.106+(0,8-x).40=41,1\\ \to x=0,7\\ \to n_{CO_2}=\dfrac{1}{2}.n_{NaOH\ \text{phản ứng}}=0,35(mol)\\ \to V_{CO_2}=0,35.22,4=7,84(l)\\ \to V=7,84$
$3/\\ 1,\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ n_{NaOH}=0,2.1=0,2(mol)\\ Đặt\ T=\dfrac{n_{CO_2}}{n_{NaOH}}=\dfrac{0,15}{0,2}=0,75\\ 0,5<T<1\to \text{Tạo 2 muối}\\ Đặt\ \begin{cases}n_{NaHCO_3}=x(mol)\\_{Na_2CO_3}=y(mol)\end{cases}\\ NaOH+CO_2\to NaHCO_3\\ 2NaOH+CO_2\to Na_2CO_3+H_2O\\ \to \begin{cases}x+y=0,15\\x+2y=0,2\end{cases}\\ \to \begin{cases}x=0,1\\y=0,05\end{cases}\\ \to m_{Muối}=0,1.84+0,05.106=13,7(g)\\ \to m=13,7$
$2,\\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ n_{KOH}=0,5.0,5=0,25(mol)\\ Đặt\ T=\dfrac{n_{CO_2}}{n_{KOH}}=\dfrac{0,1}{0,25}=0,4\\ T<0,5<1\to \text{Tạo muối trung hòa, KOH dư}\\ 2KOH+CO_2\to K_2CO_3+H_2O\\ n_{K_2CO_3}=n_{CO_2}=0,1(mol)\\ n_{KOH\ dư}=0,25-0,2=0,05(mol)\\ \to m_X=0,05.56+0,1.138=16,6(g)$
