Answer is: 33.34 milliliters of HCl.
Balanced chemical reaction: Zn(s) + 2HCl(aq) → H₂(g) + ZnCl₂(aq).
m(ore) = 5.45 g.
ω(Zn) = 32.0% ÷ 100%.
ω(Zn) = 0.32; mass percentage of zinc in an ore.
m(Zn) = ω(Zn) · m(ore).
m(Zn) = 0.32 · 5.45 g.
m(Zn) = 1.635 g.
n(Zn) = m(Zn) ÷ M(Zn).
n(Zn) = 1.635 g ÷ 65.38 g/mol.
n(Zn) = 0.025 mol; amount of the substance.
From chemical reaction: n(Zn) : n(HCl) = 1 : 2.
n(HCl) = 2 · 0.025 mol = 0.05 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0.05 mol ÷ 1.50 mol/L.
V(HCl) = 0.033 L · 1000 mL/L.
V(HCl) = 33.34 mL; volume of hydrochloric acid.