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.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i