Answer to Question #297915 in General Chemistry for Phia

Question

How many mL of 0.250 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 50 mL of 0.350 M aluminum sulfate?

3 Ba(NO3)2 (aq) + Al2(SO4)3 (aq) --> 3 BaSO4 (s) + 2 Al(NO3)3 (aq)

Answer 1

Moles of aluminium sulfate used=

(50/1000)× 0.350= 0.0175moles

Mole ratio= 3:1

Moles of barium nitrate=

0.0175× 3= 0.0525moles

(V/ 1000)× 0.250= 0.0525

V= 0.0525/0.00025 = 210ml