\(n_{HCl}=0,2\cdot0,2=0,04mol\)\(\Rightarrow n_{H^+}=0,04mol\)
Để trung hòa:\(\Rightarrow n_{OH^-}=n_{H^+}=0,04mol\)
\(\Rightarrow n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{OH^-}=\dfrac{1}{2}\cdot0,04=0,02mol\)
\(\Rightarrow m_{Ca\left(OH\right)_2}=0,02\cdot74=1,48\left(g\right)\)
\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{1,48}{25}\cdot100=5,92\left(g\right)\)