\(m_{H_2SO_4}=\dfrac{m_{ddH_2SO_4}\cdot C\%}{100\%}=\dfrac{200\cdot9,8\%}{100\%}=19,6g\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(\dfrac{0,2}{3}\) 0,2
\(m_{Al_2O_3}=\dfrac{0,2}{3}\cdot102=6,8g\)