Which of two curves exhibits exponential growth

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

Alik [6]

Answer:

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

Explanation:

This is an unrealistic problem, because Al(OH)₃is highly insoluble in water.

There are two parts to this question:

A. Stoichiometry — in which we figure out the volumes, masses, and moles of products

B. Calorimetry — in which we calculate the enthalpy of reaction.

A. Stoichiometry

1. Calculate the volume of Al(OH)₃

(a) Balanced chemical equation.

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃+ 6H₂O

V/mL: 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 is 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

There are two energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm 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_i = 22.3 °C

T_f = 24.7 °C

Calculations

(a) Mass of solution

Assume the solutions have the same density as water (unrealistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_f - Ti = 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 is an absurd answer, but it's what comes from your numbers.

6 0

3 years ago

How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the wor

tangare [24]

Answer is: 4.45 grams of methane gas need to be combusted.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V = n·R·T.
p = 1.1 atm.
T = 301 K.
V(H₂O) = 12.5 L.
R = 0,08206 L·atm/mol·K.
n(H₂O) = 1.1 atm · 12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.

4 0

3 years ago

Which chemical equation correctly represents the synthesis of calcium sulfite from calcium oxide and sulfur dioxide?

OverLord2011 [107]

The chemical equation that correctly represents the synthesis of calcium sulfite is

CaO(s) + SO₂(g) → CaSO₃(s)

Explanation

Synthesis reaction is type of chemical reaction in which two or more simple substance combine to form a more complex product.

The reaction above is therefore a synthesis reaction since CaO and So₂ combine to form CaCO₃ ( a complex product)

5 0

3 years ago

Read 2 more answers

a chemist want to make a solution of 3.4M HCl there are tow solutions of HCl that he can find on the shelf. one has a concentrat

steposvetlana [31]

Answer:

The most concentrated one, 6.0 M.

Explanation:

A simple and reliable way to produce a solution of HCl (or anything else, for that matter) is to use a more concentrated solution and dilute it.

In this case the chemist could take a portion of the 6.0 M solution and dilute it by adding water, until the concentration is 3.4 M.

Such a process would not be possible with the 2.0 N (which is the same as 2.0 M for HCl) solution.

7 0

3 years ago

QUESTION 18 Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) 3O 2 (g) 2Al 2O

Delicious77 [7]

Answer:

Percent yield = 85.4%

Explanation:

Given data:

Mass of Al = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

First of all we will calculate the number of moles of each reactant.

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with oxygen and aluminium.

Al : Al₂O₃

4 : 2

0.09 : 2/4×0.09 = 0.045

O₂ : Al₂O₃

3 : 2

0.08 : 2/4×0.08 = 0.04

The number of moles of aluminium oxide produced by oxygen are less so it will limiting reactant.

Theoretical yield of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.04 mol × 101.96 g/mol

Mass = 4.1 g

Percent yield:

Percent yield = actual yield /theoretical yield × 100

Percent yield = 3.5 g/ 4.1 g × 100

Percent yield = 85.4%

7 0

3 years ago