\(SO_3+H_2O\rightarrow H_2SO_4\\ n_{H_2SO_4\left(tổng\right)}=\dfrac{196.10\%}{98}+\dfrac{98a}{80}=0,2+\dfrac{49}{40}a\left(mol\right)\\ m_{ddH_2SO_4\left(tổng\right)}=a+196\left(g\right)\\ Vì:C\%_{ddH_2SO_4\left(cuối\right)}=20\%\\ \Leftrightarrow\dfrac{0,2+\dfrac{49}{40}a}{a+196}.100\%=20\%\\ \Leftrightarrow a\approx38,049\left(g\right)\)