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23 days ago

How many moles NaOH are contained in 65.0 mL of a 2.20 M solution of NaOH in H2O?

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66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu

eduard [2782]

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}$

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V: 66.667

c/mol·L⁻¹: 4.000 3.000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}$

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d) Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}$

This result appears nonsensical, but it is derived from your given figures.

6 0

3 months ago

Select the word or phrase from the drop-down menu to describe ionic compounds. A formula unit represents the simplest ratio of e

VMariaS [2998]

Response:

CRYSTAL

A LARGE NUMBER OF ATOMS ORGANIZED IN A REGULAR STRUCTURE

1:1

Reasoning:

8 0

2 months ago

Read 2 more answers

In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

KiRa [2933]

I predict that there will be an increase in the seconds recorded in the time column. This is because, as more water is mixed with sodium thiosulfate, its concentration diminishes in each flask. Additionally, a lower concentration results in a slower reaction rate since fewer molecules of sodium thiosulfate means there are less frequent collisions with sulfuric acid. With fewer collisions occurring in the reaction, it takes a longer time for the reaction to complete, leading to increased time when sodium thiosulfate is diluted.

Explanation:

I can confirm that this explanation is accurate.

6 0

3 months ago