Answer to Question #319200 in General Chemistry for Emman

Question

To precipitate all the sulfate ions from 22.0 mL of 0.150 molarity aluminum sulfate as barium sulfate, how many mL of 0.250 molarity3. barium nitrate are required? (8 points)

____Ba(NO3)2(aq)+____Al2(SO4)3(aq)→______BaSO4(s)+____Al(NO3)3(aq)

Answer 1

2Ba(NO3)2 + Al2(SO4)3 = 3Ba(SO4) + 2Al(NO3)3

n(Al2(SO4)3) = M*V = 0.150*0.022 = 0.0033 mol

n(Ba(NO3)2) = 0.0033*3 = 0.0099 mol

V(Ba(NO3)2) = n/M = 0.0099/.0250 = 0.0396l (39.6 ml)

Answer 39.6ml