Equation given and Is balanced The Chemistry of Recycling Percent Yield: The fin
ID: 1029883 • Letter: E
Question
Equation given and Is balancedThe 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 %
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