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?

Question

Chemistry

Shaylee Peterson

7 September, 03:46

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?

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Sloane Ali · Answer 1

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.