Question

Cryolite, Na3AlF6, is an important industrial reagent. It is made by the reaction below.
Al2O3(s)+6NaOH(aq)+12HF(g)→2Na3AlF6(s)+9H2O(l)
In an experiment, 7.81 gAl2O3 and excess HF(g) were dissolved in 3.50 L of 0.141 MNaOH. If 25.1 g Na3AlF6 was obtained, then what is the percent yield for this experiment?

Answer 1

moles of NaOH = 3.50 x 0.141 = 0.4935

according to balanced reaction

101.96 g AL2O3 reacts with 6 moles NaOH

7.81 g Al2O3 reacts with 7.81 x 6 / 101.96 = 0.46 moles NaOH

but we have 0.4935 moles NaOH means exess NaOH present

Al2O3 is limiting reagent.

101.96 g Al2O3 gives 2 x 209.94 g Na3AlF6

7.81 g Al2O3 gives 7.81 x 2 x 209.94 / 101.96 = 32.16 g Na3AlF6

therotical yield = 32.16 g

actual yield = 25.1 g

% yield = (actual yield /therotical yield) x 100

% yield = (25.1 / 32.16) x 100

% yield = 78.05