Question

Alum was synthesized in the same manner as described in this experiment. If the actual yield of alum was 0.8675 g and the percent yield was 72.0%, what mass of aluminum was initially used? 2 Al(s) + 2 KOH(as) + 4 H2SO4(aq) + 22 H2O(l)  2 [KAl (SO4)2 ∙ 12 H2O](s) + 3 H2(g)

Answer 1

Molar mass of Al = 27 g/mol

Molar mass of Alum KAl(SO4)2 ∙ 12 H2O = 474 g/mol

Given that percent yield = 72 %

percent yield = (actual yield / theoretical yield) x100

72 = (0.8675 g/   theoretical yield) x 100

Then,

   theoretical yield = 1.2 g

Hence,

2 Al(s) + 2 KOH(as) + 4 H2SO4(aq) + 22 H2O(l) --------> 2 [KAl(SO4)2 ∙ 12 H2O](s) + 3 H2(g)

2 mol 2 mol

1 mol 1 mol

27 g 474 g

? 1.2 g

? = (27g ) x (1.2 g/ 474 g)

= 0.068 g

Therefore,

mass of aluminum was initially used = 0.068 g