Question

A chemistry student wants to synthesize alum (KAl(SO4)2.12H2O) from 12 g of potassium hydroxide (KOH) and excess sulfuric acid (H2SO4). If the actual yield of the reaction is 9.3 g, what is the percent yield of this reaction?

Answer 1

To calculate the percent yield, we first need to determine the theoretical yield of alum from the given amount of potassium hydroxide.1. Calculate the moles of potassium hydroxide  KOH :KOH molar mass = 39.1  K  + 15.999  O  + 1.008  H  = 56.107 g/molmoles of KOH =  12 g  /  56.107 g/mol  = 0.2139 mol2. Write the balanced chemical equation for the synthesis of alum:2 KOH + Al2 SO4 3 + 12 H2O  2 KAl SO4 212H2OFrom the balanced equation, we can see that 2 moles of KOH react to produce 2 moles of alum. Therefore, the mole ratio of KOH to alum is 1:1.3. Calculate the moles of alum  KAl SO4 212H2O  produced:Since the mole ratio is 1:1, the moles of alum produced are equal to the moles of KOH used: 0.2139 mol.4. Calculate the theoretical yield of alum:Alum molar mass = 39.1  K  + 26.98  Al  +  4  32.07   S  +  24  15.999   O  +  24  1.008   H  = 474.4 g/moltheoretical yield =  0.2139 mol    474.4 g/mol  = 101.5 g5. Calculate the percent yield:percent yield =  actual yield / theoretical yield   100percent yield =  9.3 g / 101.5 g   100 = 9.16%The percent yield of this reaction is 9.16%.