Question

Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 6.50 M HCl(aq) are required to react with 4.45 g of an ore containing 39.0 % Zn(s) by mass?

Answer 1

Mass of an in ore = 39 x 4.45 100 = 1.735g moles of Zn = 1.735 g = 0.0265 mol 65.38 g/mol Zuls) + 2Hchlag 2n C2(aq) +H219) Indl zu react with and Hel moles of He= ax0.0265 = 0.053 mot mass molarity & moles volume nounce = 0.053 8.15.103 L 6.50 Volume = 8.15mL