Question

What is the experimental yield (in g of precipitate) when 18.5 mL of a 0.632 M solution of barium hydroxide is combined with 17.8 mL of a 0.733 M solution of aluminum nitrate at a 78.6% yield?

Answer 1

Solution- The equation is

3 Ba(OH)2 + 2 Al(NO3)3 → 2 Al(OH)3(s) + 3 Ba(NO3)2

(18.5 mL) * (0.632 M Ba(OH)2) = 11.692 mmol Ba(OH)2
(17.8 mL) x (0.733 M Al(NO3)3) = 13.047 mmol Al(NO3)3

Now 11.692 m moles of Ba(OH)2 would react completely with 11.692 x (2/3) = 7.794​ m moles of Al(NO3)3 but there is more Al(NO3)3 present than that, Hence Al(NO3)3 is in excess and Ba(OH)2 is the limiting reactant.

(11.692 mmol Ba(OH)2) x (2/3) x (78.00367 g Al(OH)3/mol) x (0.786)

= 477.89 mg

= 0.478 g Al(OH)3