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Equation given and Is balanced The Chemistry of Recycling Percent Yield: The fin

ID: 1029883 • Letter: E

Question

Equation given and Is balanced
The Chemistry of Recycling Percent Yield: The final step is to determine the yield of the reaction, both in terms of the mass of the final product obtained and as the percent yield, which indicates the efectiveness of the reaction. Mass of watch glass + alum 7D-404 Mass of watch glasSk.0 Mass of alum recovered 3,4115 Since the formula mass of aluminum and alum are often very different, you cannot compare the mass of product obtained against the mass of starting material used in the reaction to determine the theoretical and yield of a reaction. To do this, you need to know how many moles of aluminum go into each mole of alum (Hint: Look at the overall reaction!) and the formula weight of alum. Summarize your results in the spaces below, and show all calculations in the spaces where indicated. percent Moles aluminum : moles alum Formula weight of alum Theoretical yield alum Percent yield alum Calculation of Formula Weight of Alum, KAIlSO2- 12 H,0 g/mol Calculation of Theoretical Yield Percent Yield Calculation

Explanation / Answer

1) Moles of Aluminum; moles of alum = 1:1

2) Calculation of formulla weight of Alum KAl(SO4)2. 12 H2O = 474.2095

Explaination: Molar Mass of K = 39.0986

Molar Mass of Al = 26.981539

Molar Mass of S = 32.065

  Molar Mass of O = 16

  Molar Mass H2O = 18

KAl(SO4)2. 12 H2O = 39.0983 + 26.981539 + (32.065X2) + (16X8) + (12X18) = 474.2095

3) Theoretical yield of Alum = 17.575333

Explaination: Moles of Aluminium = moles of Alum

Molar mass of Aluminium = molar Mass of Alum

Aluminium = Alum

26.981539 gm = 474.2095 gm

1 gm = X

   X= 17.575333

4) Percent Yeild of Alum = 79.4949 %

Explaination: 17.575333 gm of alum = 100 % yield

13.9715 gm of alum = X % yield

X = 13.9715X100/17.575333

   X= 79.4949 %