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ruslelena [56]
3 years ago
15

50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calorimeter. The reaction mixture is initially at 22.4

degree C, and the final temperature after reaction is 30.2 degree C. What is the molarity of the HCl solution? You may assume that there is an excess of base (so that all the HCl has reacted), that the specific heat of the reaction mixture is 0.96 cal/g degree C, and that the density of the reaction mixture is 1.02 g/mL. The heat of neutralization of HC1 and NaOH is 13.6 kcal/mole.
Chemistry
1 answer:
VikaD [51]3 years ago
3 0

Answer:

2.1 M is the molarity of the HCl solution.

Explanation:

HCl+NaOH\rightarrow H_2O+NaCl

Molarity of HCl solution = M_1=?

Volume of HCl solution = V_1=50.0mL

Ionizable hydrogen ions in HCl = n_1=1

Molarity of NaOH solution = M_2=1.05 M

Volume of NaOH solution = V_2=100.0 mL

Ionizable hydroxide ions in NaOH = n_2=1

n_1M_1V_1=n_2M_2V_2 (neutralization )

M_1=\frac{M_2V_2}{V_1}=\frac{1.05M\times 100.0 mL}{50.0 mL}

M_1=2.1 M

2.1 M is the molarity of the HCl solution.

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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
ollegr [7]

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

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