Question

Calculate the number of milliliters of 0.433 M Ba(OH)₂ required to precipitate all of the Fe³⁺ ions in 156 mL of 0.487 M Fe₂(SO₄)₃ solution as Fe(OH)₃. The equation for the reaction is:

Fe₂(SO₄)₃(aq) + 3Ba(OH)₂(aq) 2Fe(OH)₃(s) + 3BaSO₄(aq)

_______ mL Ba(OH)₂

Answer 1

Here:

M(Fe2(SO4)3)=0.487 M

M(Ba(OH)2)=0.433 M

V(Fe2(SO4)3)=156.0 mL

According to balanced reaction:

3*number of mol of Fe2(SO4)3 =1*number of mol of Ba(OH)2

3*M(Fe2(SO4)3)*V(Fe2(SO4)3) =1*M(Ba(OH)2)*V(Ba(OH)2)

3*0.487 M *156.0 mL = 1*0.433M *V(Ba(OH)2)

V(Ba(OH)2) = 526 mL

Answer: 526 mL