Question

Assume a student used 2.27g of aluminum metal to synthesize alum. What would be this students theoretical yield?

If the same stufent recovered 7.89g of alum, what is the percent yield?

Answer 1

We know, number of mole = Mass Molari mass Griven, mass of Aluminium metal = 2.27 g Again, moloni mass of Aluminium, AL = 26.981539 g/mol Therefore, number of mole of Aluminium (Al) = 2.27 g 26.981539 glmol = 0.084 mol balanced The chemical ecuation for the synthenize of Blum From Aluminium metal is, → 2* [Al(S04), J.12H20+ 3 H2 2 A8 + 2KOH + 4 H 904 + 22 H2O (Alum) From the above balanced chemical earuation, we can write 2 moles of Aluminium producen 2 molen of Alum 2 mol Then, 0.084 mol of Aluminium Producer = X0.084 mol 2 mol of Alum mol of Alum 20.084 2 0.168 2 = mol of Alim = 0.084 mol of Alum

= (0.084 mol) (474.3884 g/mol) of Alum (AB, molar mass of k [Al(S04).J.12H0 = 474.3884 g/mol) 39.85 g of Alum From 2. 27 g of Aluminium, 39.85g of Alum is produced which is ište Theoretical yield of Alum. So, theoretical yield eld of Alum is 39.85 g Now, percent yield Actual yield Theoretical yield Given, The estudent recovered 7.89 g of Alum. Hence, actual yield of Alum = 7.899 Again, theoretical yield of Alum = 39.859 Actual yield Now, Percent yield = Theoretical yield 7.899 39.85g = X 100%, X 100 X 100% 0.198 X 100 % = 19.8% Therefore, percent yield of Alum obtained ins 19.8.1,