\(NaOH+HCl--->NaCl+H2O\)
\(Na2CO3+2HCl--.2NaCl+H2O+CO2\uparrow\)
\(n_{CO2}=\frac{4,48}{22,4}=0,1\left(mol\right)\)
\(n_{Na2CO3}=n_{CO2}=0,2\left(mol\right)\)
\(n_{Na2CO3}=106.0,2=21,2\left(mol\right)\)
\(\%\%m_{Na2CO3}=\frac{21,2}{30}.100\%=70,67\%\)