(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.
Because while doing research scientist follow the scientific metod.The scientific method involves exploration and observation and the search for the answer for a question. And there will always some hypotesis that while experimenting reveal to be wrong.
I think it will be B. Hope it helps :)
