\(m_{NaOH}=\dfrac{200\cdot10\%}{100\%}=20\left(g\right)\\ \Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\\ PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,25\cdot98=24,5\left(g\right)\)