Question

Zinc reacts with hydrochloric acid according to the reaction equation

Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)

How many milliliters of 2.00 M HCl(aq)2.00 M HCl(aq) are required to react with 8.55 g Zn(s)?

Answer 1

Balanced chemical equation is:

Zn + 2 HCl ---> ZnCl2 + H2O

Molar mass of Zn = 65.38 g/mol

mass(Zn)= 8.55 g

use:

number of mol of Zn,

n = mass of Zn/molar mass of Zn

=(8.55 g)/(65.38 g/mol)

= 0.1308 mol

According to balanced equation

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

= (2/1)*0.1308

= 0.2615 mol

This is number of moles of HCl

use:

M = number of mol / volume in L

2.0 = 0.2615/ volume in L

volume = 0.1308 L

volume = 1.308*10^2 mL

Answer: 131 mL