How many milliliters of 1.50 m hcl(aq) are required to react with 5.45 g of an ore containing 32.0% zn(s) by mass?

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.

the product of the pressure and volume is also constant