Question

What is the experimental yield (in g of precipitate) when 19.8 mL of a 0.663 M solution of barium hydroxide is combined with 19.2 mL of a 0.727 M solution of aluminum nitrate at a 77.4% yield?

Answer 1

volume of Ba(OH)2, V = 19.8 mL

= 1.98*10^-2 L

use:

number of mol in Ba(OH)2,

n = Molarity * Volume

= 0.663*1.98*10^-2

= 1.313*10^-2 mol

volume of Al(NO3)3, V = 19.2 mL

= 1.92*10^-2 L

use:

number of mol in Al(NO3)3,

n = Molarity * Volume

= 0.727*1.92*10^-2

= 1.396*10^-2 mol

Balanced chemical equation is:

3 Ba(OH)2 + 2 Al(NO3)3 ---> 2 Al(OH)3 + 3 Ba(NO3)2

3 mol of Ba(OH)2 reacts with 2 mol of Al(NO3)3

for 1.313*10^-2 mol of Ba(OH)2, 8.752*10^-3 mol of Al(NO3)3 is required

But we have 1.396*10^-2 mol of Al(NO3)3

so, Ba(OH)2 is limiting reagent

we will use Ba(OH)2 in further calculation

Molar mass of Al(OH)3,

MM = 1*MM(Al) + 3*MM(O) + 3*MM(H)

= 1*26.98 + 3*16.0 + 3*1.008

= 78.004 g/mol

According to balanced equation

mol of Al(OH)3 formed = (2/3)* moles of Ba(OH)2

= (2/3)*1.313*10^-2

= 8.752*10^-3 mol

use:

mass of Al(OH)3 = number of mol * molar mass

= 8.752*10^-3*78

= 0.6827 g

% yield = actual mass*100/theoretical mass

77.4= actual mass*100/0.6827

actual mass=0.5284 g

Answer: 0.528 g