2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 10 H2O(l) ⟶ 2 KAl(SO4)2∙12 H2O(s) + 3 H2(g)  alum  Using the data, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant. Description mass bottle mas 9.150 bottle mass with aluminum pieces 11.177 final product and bottle masss 19.377 What is the theoretical yield of alum and what is the percent yield of alum?

If 0.315 g of AI (26.982 g/mol) is reacted in the synthesis of Alum reaction and 3.265 g of Alum (474.39 g/mol) is produced, determine the percent yield. O 60% 58% O 57% O 55% O 59% O 56%

2 N₂O5(g) --> 4 NO₂(g) + O₂(g) True/False: The equation is already balanced as it is written.

For the experiment " limiting reactant " and after filtering the precipitate, you checked the filtrate (supernatant) to identify the limiting reactant. 2Na3PO4 + 3BaCl2 → Ba3(PO4)2 (s) + 6NACI If you added BaCl2 to the filtrate (supernatant) and a white precipitate was formed, this means that: O a. None of the mentioned O b. BaCl2 is an excess Oc. Both BaC2 and Na3PO4 are limiting reactants O d. BaCl2 is the limiting reactant O e. NagPO4 is the limiting reactant