Question

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Experiment 2 - Preparation of Alum Evaluation Question 1 Part (a) Calculate the volume of 1.0 M KOH solution required to completely dissolve 0.5589 g of aluminium. The balanced chemi follows: 2Al(s) + 2KOH(aq) + 6H2O(0) -> 2Al(OH)4 (aq) + 2K+ (aq) + 3H2(g) (Please give your answer to 2 significant figures.) mL KOH Check Answer Part (b) The 0.5589 g of aluminium metal from Part (a), reacted with excess potassium hydroxide to produce 3.801 g of alum. The balanced chemical equation for the reaction is as follows: AP+ (aq) + K+ (aq) + 2504² (aq) + 12H2O(1) -> KAI(SO4)2 . 12H20 (9) Determine the percentage yield for the reaction. (Please give your answer to 4 significant figures) 7% Check Answer

Answer 1

Part A

from balanced equation:

   2 mol Al = 2 mol KOH

No of mol of Al reacted = 0.5589/27 = 0.0207 mol

No of mol of KOH required = 0.0207 mol

volume of KOH required = 0.0207/1 = 0.0207 L

                                  = 20.7 ml

Part B

from the above,

No of mol of Al reacted = 0.5589/27 = 0.0207 mol

No of mol of Alum produced = 0.0207 mol

massof alum formed = 0.0207*258 = 5.34 g

percent yield = experimental yield/theoretical yield*100

              = 3.801/5.34*100

              = 71.2%