(10.0 g Al2(SO3)3) / (294.1544 g Al2(SO3)3/mol) = 0.033996 mol Al2(SO3)3 (10.0 g NaOH) / (39.99715 g NaOH/mol) = 0.25002 mol NaOH
0.033996 mole of Al2(SO3)3 would react completely with 0.033996 x (6/1) = 0.203976 mole of NaOH, but there is more NaOH present than that, so NaOH is in excess and Al2(SO3)3 is the limiting reactant.
