2NaOH + H2SO4 → Na2SO4 + 2H2O
\(m_{NaOH}=200\times10\%=20\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,5=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25\times98=24,5\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)