Question

Zinc reacts with hydrochloric acid according to the reaction equation Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H,(8) How many milliliters of 2.50 M HCl(aq) are required to react with 2.75 g Zn(s)? volume:

Answer 1

2.75 g zu mol of Zn = man molat man 2.75 g 65.38 g/mol = 0.04206 mol Zn (6) Zn (3) + 2HCl caa) - Zilla (ar) + H₂ cg, mol Ratio of HCl and Zn = 2:1 moles of HCl moles of an = 2 moles of Hel needed = (moles of Zn) x2 = 0.04206 x 22 = 0.08412 mol HCl needed we have 2.50 M Hol Solution molarity = moles of Hol Volume in litre litre volume = volume 0.08412 2.50 mele lor moles of HCl = molarity Volume 0.033648 litre . Volume = 33.648 mL = 33.6 mL