\(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(0.1................................0.1\)
\(m_{dd}=10.2+300=310.2\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0.1\cdot342}{310.2}\cdot100\%=11.02\%\)
