Đáp án:

 50g và 200g

Giải thích các bước giải:

\(\begin{array}{l}
{C_\% }{H_2}O\, = 0\% \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,100 - 20 = 80\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \searrow \,\,\,\,\,\,\,\,\,\, \nearrow \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,20\% \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \nearrow \,\,\,\,\,\,\,\,\, \searrow \\
{C_\% }BaC{l_2} = 100\% \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,20 - 0 = 20\\
\frac{{{m_{{H_2}O}}}}{{{m_{BaC{l_2}}}}} = \dfrac{{80}}{{20}} = 4 \Rightarrow 4{m_{BaC{l_2}}} = {m_{{H_2}O}}\\
{m_{{\rm{dd}}}} = 250g \Rightarrow {m_{BaC{l_2}}} + {m_{{H_2}O}} = 250g\\
 \Rightarrow 4{m_{BaC{l_2}}} + {m_{BaC{l_2}}} = 250 \Rightarrow {m_{BaC{l_2}}} = 50g\\
{m_{{H_2}O}} = 4 \times 50 = 200g
\end{array}\)