Ta có: \(n_{H_2SO_4}=\dfrac{\dfrac{9,8\%.200}{100\%}}{98}=0,2\left(mol\right)\)
\(PTHH:2NaOH+H_2SO_4--->Na_2SO_4+2H_2O\)
Theo PT: \(n_{NaOH}=2.n_{H_2SO_4}=2.0,2=0,4\left(mol\right)\)
\(\Rightarrow V_{dd_{NaOH}}=\dfrac{0,4}{1}=0,4\left(lít\right)=400\left(ml\right)\)
Chọn D