Question

What is the experimental yield (in g of precipitate) when 17.9 mL of a 0.54 M solution of barium hydroxide is combined with 15.4 mL of a 0.674 M solution of aluminum nitrate at a 87.6% yield?

Answer 1

2 Al(NO3)3(aq) + 3 Ba(OH)2(aq) = 2 Al(OH)3(s) + 3 Ba(NO3)2(aq)

17.9 ml of 0.54 M barium hydroxide combined with 15.4 ml of 0.674 M aluminium nitrate

acording to balence chemical equation 2 mole of aluminium nitrate react with 3 mole of barium hydroxide

number of moles of barium hydroxide = 0.54*0.0179 L= 0.009666 moles

and number of moles of aluminium nitrate = 0.674*0.0154 L = 0.01037 moles

limiting ragent is barium hydroxide so

theoritical yield = (0.009666*2) / 3= 0.006444 moles

(experimental / theoritical)*100 = 87.6 %

experimental yield = 0.005644 moles

experimental yield in g =(0.005644 * 78g/mol) = 0.44g