Question

Zinc dissolves in hydrochloric acid to yield hydrogen gas: Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g) When a 12.7 g chunk of zinc dissolves in 5.00 x 102 mL of 1.450 M HCl, what is the concentration of hydrogen ions remaining in the final solution?0 M0.388 M0.674 M0.776 M1.06 M

Answer 1

Answer:

$\boxed{\text{0.673 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Data:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:            65.38

                     Zn   +   2HCl ⟶ ZnCl₂ + H₂

m/g:             12.7

V/mL:                      5.00×10²

c/mol·L⁻¹:                  1.450

2. Moles of each reactant  

(a) Moles of Zn

$n = \text{12.7 g Zn} \times \dfrac{\text{1 mol Zn}}{\text{65.38 g Zn}} = \text{0.1942 mol Zn}$

(b) Moles of HCl

V = 5.0× 10² mL = 0.5000 L

$n = \text{0.5000 L HCl}\times \dfrac{\text{1.450 mol HCl}}{\text{1 L HCl}} = \text{0.7250 mol HCl}$

3. Identify the limiting reactant

Calculate the moles of ZnCl₂ obtained from each reactant

(i) From Zn

The molar ratio is 1 mol ZnCl₂:1 mol Zn

$n = \text{0.1942 mol Zn} \times \dfrac{\text{1 mol ZnCl}_{2}}{\text{1 mol Zn}} = \text{0.1942 mol ZnCl}_{2}$

(ii) From HCl

The molar ratio is 1 mol ZnCl₂:2 mol HCl

$n = \text{0.7250 mol HCl} \times \dfrac{\text{1 mol ZnCl}_{2}}{\text{2 mol HCl}} = \text{0.3625 mol ZnCl}_{2}$

Zinc is the limiting reactant, because it produces fewer moles of ZnCl₂.

4. Moles of HCl reacted

The molar ratio is 2 mol HCl:1 mol Zn

$n = \text{0.1942 mol Zn} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Zn}} = \boxed{\text{0.3885 mol HCl}}$

5. Moles of HCl remaining

n = 0.7250 - 0.3885 = 0.3365 mol HCl

6. Concentration of hydrogen ions

The HCl is completely dissociated.

$c = \dfrac{\text{0.3365 mol}}{\text{0.5000 L}} = \textbf{0.673 mol/L}\\\\\text{The concentration of hydrogen ions is \boxed{\textbf{0.673 mol/L}} }$