Question

A student started a synthesis of alum with 3.4x10^-2 mole aluminum:

2Al (s) + 6h2O (I) + 2KOH (aq)------> 2K[Al(OH)4] (aq) + 3H2 (g) (1)

2K[Al(OH)4] (aq) + H2SO4 (aq) ------> 2Al(OH)3 (s) +2 K2SO4 (aq) + 2H2O (I) (2)

2Al(OH)3 (s) + 3H2SO4 (aq) ---------> Al2(SO4)3 (aq) + 6H2O (3)

Al2(SO4)3 (aq) + K2SO4 (aq) + 24 H2O---------> 2KAl(SO4)2 12H2O (4)

a) What theoretical yield of alum, in gram, would be produced from 3.4x10^-2 mole of aluminum?

b) A student obtained 12.4 g alum from the synthesis. If the theortical yield was 19.6 g, what was the percent yield?

Answer 1

Ans. a. From the stoichiometry of equation 1 through 4-

2 moles of aluminum produce 2 moles of alum.

Thus, 3.4 x 10^-2 moles of aluminum produce 3.4 x 10^-2 moles of alum.

Molecular mass of alum = 474.39 gm/ mol

Theoretical yield of alum (g) = moles of alum x molecular mass of alum

                                           =3.4 x 10^-2 moles x 474.39 gm/ mol    

                                           = 16.13 gm

Ans. b. % yield = (yield obtained / theoretical yield) x 100

                        = (12.4 gm / 19.6 gm) x 100

                        = 63.27 %