Question

What is the experimental yield (in g of precipitate) when 16.5 mL of a 0.633 M solution of barium hydroxide is combined with 17.3 mL of a 0.521 M solution of aluminum nitrate at a 89.6% yield?

What is the theoretical yield (in g of precipitate) when 16.4 mL of a 0.559 M solution of iron(III) chloride is combined with 16.9 mL of a 0.577 M solution of lead(II) nitrate?

What is the theoretical yield (in g of precipitate) when 18.4 mL of a 0.73 M solution of sodium phosphate is combined with 18 mL of a 0.695 M solution of aluminum chloride?

Answer 1

1)

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

Now, in the reaction,

(16. 5mL) x (0.633 M Ba(OH)2) / 3 = 3.48mmol Ba(OH)2

(17.3mL) x (0.521M Al(NO3)3) / 2 = 4.50 mmol Al(NO3)3

Mol of Ba(OH)2 are the least so, it would be the limiting reagent.

Experimental yield = mol of limiting reagent x (stoichiometric coefficient of product / stoichiometric coefficient of Limiting reagent) x molar mass of product x (% yield)

(16.5 x 0.633 mmol Ba(OH)2) x (2/3) x (78.00367 g Al(OH)3/mol) x (0.893)

= 486 mg

= 0.486 g Al(OH)3