a) H2SO4 + 2KOH → K2SO4 + 2H2O
b) \(m_{H_2SO_4}=200\times9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,4\times56=22,4\left(g\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{22,4}{8\%}=280\left(g\right)\)