Zinc reacts with hydrochloric acid according to the reaction equation shown. zn(s) + 2 hcl(aq) → zncl2(aq) + h2(g) how many milliliters of 2.50 m hcl(aq) are required to react with 4.65 g of an ore containing 50.0% zn(s) by mass? volume: ml

Volume of HCl required = 28.4 mL

Explanation:

Mass of zinc ore = 4.65 g

% of zinc in zinc ore = 50 %

So, mass of zinc in zinc ore = 0.50 × 4.65 = 2.325 g

No. of moles of Zn =

So, as per the reaction coefficient,

1 mol of zinc reacts with 2 mol of HCl

0.0355 mol of zinc reacts with

                                    = 0.0355 × 2 = 0.071 mol of HCl

Molarity of HCl = 2.50 M

Volume of HCl =

1 L = 1000 mL

0.0284 L = 1000 × 0.0284 = 28.4 mL

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