a) $Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
b) $n_{Al_2(SO_4)_3} = n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$
$m_{Al_2(SO_4)_3} = 0,1.342 = 34,2(gam)$$
c) $m_{dd\ sau\ pư} = 10,2 + 200 = 210,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{34,2}{210,2}.100\% = 16,27\%$