What volume of 0.030 M Fe3+ solution is required to react with 77.0 mL of 0.158 M OH– solution in a precipitation titration?

Balanced equation for the precipitation:
Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)
Mole ratio Fe³⁺ : OH⁻ = 1 : 3

Moles of OH⁻ reacted = (0.158 mol/L) × (77.0/1000 L) = 0.012166 mol
Moles of Fe³⁺ needed = (0.012166 mol) × (1/3) = 0.004055 mol
Volume of Fe³⁺ solution = (0.004055 mol) / (0.030 mol/L) = 0.135 L = 135 mL

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OR:
(0.158 mol OH⁻ / 1000 mL OH⁻ solution) × (77.0 mL OH⁻ solution) × (1 mol Fe³⁺ / 3 mol OH⁻) × (1000 mL Fe³⁺ solution / 0.030 mol Fe³⁺)
= 135 mL Fe³⁺ solution