\(\begin{array}{l}
160ml=0,16l\\
n_{CO_2}=\frac{4,48}{22,4}=0,2(mol)\\
n_{Ba(OH)_2}=0,16.1=0,16(mol)\\
\to T=\frac{0,16}{0,2}=0,8\\
\to 0,5<T<1\\
\to Tao\,2\,muo\,BaCO_3:x(mol);Ba(HCO_3)_2:y(mol)\\
\rm Bảo\,toàn\,ngto\,ta\,có:\\
\begin{cases} x+y=0,16\\ x+2y=0,2 \end{cases}\to\begin{cases} x=0,12\\ y=0,04 \end{cases}\\
\to m_{muoi\,khan}=0,12.197+0,04.259=34(g)
\end{array}\)