Question

Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 4.00 M HCl(aq) are required to react with 7.15 g of an ore containing 43.0 % Zn(s) by mass?

Answer 1

Mass of Zn in ore = 43.0 % of 7.15 g

= 43.0 * 7.15 / 100

= 3.07 g

Balanced chemical equation is:

Zn + 2 HCl ---> ZnCl2 + H2O

Molar mass of Zn = 65.38 g/mol

mass(Zn)= 3.07 g

use:

number of mol of Zn,

n = mass of Zn/molar mass of Zn

=(3.07 g)/(65.38 g/mol)

= 4.696*10^-2 mol

According to balanced equation

mol of HCl reacted = (2/1)* moles of Zn

= (2/1)*4.696*10^-2

= 9.391*10^-2 mol

This is number of moles of HCl

use:

M = number of mol / volume in L

4.0 = 9.391*10^-2/ volume in L

volume = 0.02348 L

volume = 23.48 mL

Answer: 23.5 mL