What volume of 0.45 m lioh would be needed to neutralize 60.0 ml of 0.15 m hi?

Answer is: volume of 0.45 M LiOH is 20 mL.
Chemical reaction: LiOH + HI → LiI + H₂O.
From chemical reaction: n(LiOH) : n(HI) = 1 : 1.
c(LiOH) · V(LiOH) = c(HI) · V(HI).
0.45 M · V(LiOH) = 0.15 M · 60.0 mL.
V(LiOH) = 0.15 M · 60.0 mL / 0.45 M.
V(LiOH) = 20 mL ÷ 1000 mL/L = 0.02 L.
n - amount of the substance.
c - concentration of solution.

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The correct answer is c