2NaOH + H2SO4 → Na2SO4 + 2H2O
\(m_{NaOH}=140\times40\%=56\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{56}{40}=1,4\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times1,4=0,7\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,7\times98=68,6\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{68,6}{20\%}=343\left(g\right)\)
Theo pT: \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times1,4=0,7\left(mol\right)\)
\(\Rightarrow m_{Na_2SO_4}=0,7\times142=99,4\left(g\right)\)