Question

Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.00 M HCl(aq) are required to react with 2.65 g of an ore containing 26.0 % Zn(s) by mass? volume: mL

Answer 1

Write the reacti on as follows: ZnCl2 (aq)+H (g) Zn (s)+2HCI (aq) 1mole of Zn reacts with 2 moles of HCl gives 1mole of ZnCl, and 1 mole of H. Given 2.65 g mass of ore mass% of Zn in the ore = 26.0% Calculate the mass of Zn in the ore as follows mass of Zn = mass of ore xmass% of Zn 2.65 gx26.0% 26.0 2.65 gx- 100 = 0.689 g Calculate the moles of Zn in the ore as follow molar mass of Zn 65.38 g/mol mass of Zn moles of Zn molar mass of Zn 0.689 g 65.38 g/mol = 0.0105mol Molarity of HCI = 5.00 M volume of HCl = ? from the reacti on moles of Zn moles of HCl 1 2 Molarity of HClx volume of HCl moles of Zn 1 2 Molarity of HClxvolume of HCl 2xmoles of Zn 5.00 M xvolume of HCl 2x0.0105mol 0.021 mol 1000 mL volume of HC1 X X 5.00 M mol/L 1L 4.2 mL Therefore, the required soluti on is 4.2 mL