PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{H_2SO_4}=0,1\cdot0,1=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,02\left(mol\right)\\n_{Na_2SO_4}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_4}=0,01\cdot142=1,42\left(g\right)\\C\%_{NaOH}=\dfrac{0,02\cdot40}{200}=0,4\%\end{matrix}\right.\)