Em tham khảo nha:
\(\begin{array}{l}
TN1:\\
HCl + N{a_2}C{O_3} \to NaHC{O_3} + NaCl(1)\\
NaHC{O_3} + HCl \to NaCl + C{O_2} + {H_2}O(2)\\
\dfrac{{0,5}}{1} > \dfrac{{0,3}}{1} \Rightarrow\text{ HCl dư} \\
nHCl(2) = 0,5 - 0,3 = 0,2(mol)\\
nNaHC{O_3}(1) = nN{a_2}C{O_3} = 0,3\,mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,3}}{1} \Rightarrow\text{ $NaHCO_3$ dư} \\
nC{O_2} = nHCl(2) = 0,2\,mol\\
VC{O_2} = 0,2 \times 22,4 = 4,48l\\
TN2:\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
\dfrac{{0,3}}{1} > \dfrac{{0,5}}{2} \Rightarrow\text{$Na_2CO_3$ dư} \\
nC{O_2} = \dfrac{{0,5}}{2} = 0,25\,mol\\
VC{O_2} = 0,25 \times 22,4 = 5,6l
\end{array}\)