h) Find Moles of Reaction (Mg is the limiting reagent, so calc how many moles of Mg reacted from grams of Mg) 9rxn O. 453 ng 0.23 25g * /mol mg 2.168.3- 3 i) AH = qrxn /moles 240319 mg 0.0095 39 mb) Mga y counts A -06453 kJ A5111 spl) + past moles: 0.009564 mol Mg 4 AHrxn2-4104 kj/mol =-47.36 kj/mol 1.76
h) Find Moles of Reaction (Mg is the limiting reagent, so calc how many moles of Mg reacted from grams of Mg) 9rxn O. 453 ng 0.23 25g * /mol mg 2.168.3- 3 i) AH = qrxn /moles 240319 mg 0.0095 39 mb) Mga y counts A -06453 kJ A5111 spl) + past moles: 0.009564 mol Mg 4 AHrxn2-4104 kj/mol =-47.36 kj/mol 1.76
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Section h: Finding Moles of Reaction**
To find moles of reaction, it is stated that Magnesium (Mg) is the limiting reagent. The calculation is performed to find how many moles of Mg reacted from the given grams of Mg.
Given data:
- Mass of Mg = 0.23259 grams
- Molar mass of Mg = 24.31 grams/mol
Calculation:
\[
\text{Moles of Mg} = 0.23259 \, \text{g} \times \frac{1 \, \text{mol Mg}}{24.31 \, \text{g Mg}} = 0.009579 \, \text{mol Mg}
\]
The final value of moles:
\[
\text{Moles} = 0.009564 \, \text{mol Mg}
\]
**Section i: Calculating \(\Delta H\)**
The enthalpy change (\(\Delta H\)) is calculated using the heat energy released or absorbed (\(q_{\text{rxn}}\)) and the moles of reactant.
Given data:
- \(q_{\text{rxn}} = 0.453 \, \text{kJ}\)
- Moles of Mg calculated earlier = 0.009564 mol
Formula:
\[
\Delta H = \frac{q_{\text{rxn}}}{\text{moles}}
\]
Calculation for \(\Delta H\):
\[
\Delta H = \frac{-0.453 \, \text{kJ}}{0.009564 \, \text{mol Mg}} = -47.36 \, \text{kJ/mol}
\]
An additional provided value:
\[
\Delta H_{\text{rxn2}} = 47.4 \, \text{kJ/mol}
\]
Note: The calculation detail indicates a negative sign associated with \(\Delta H\), signifying an exothermic reaction.
**Conclusion:**
These calculations are critical in determining the enthalpy change associated with a chemical reaction where Mg acts as the limiting reagent. This understanding is vital for experiments requiring precise thermal measurements.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facf8e957-4fbe-471c-af8b-521f0436b36d%2F4d9d6182-74a4-4a00-81f8-fff6de475f6e%2Fet4ja8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Section h: Finding Moles of Reaction**
To find moles of reaction, it is stated that Magnesium (Mg) is the limiting reagent. The calculation is performed to find how many moles of Mg reacted from the given grams of Mg.
Given data:
- Mass of Mg = 0.23259 grams
- Molar mass of Mg = 24.31 grams/mol
Calculation:
\[
\text{Moles of Mg} = 0.23259 \, \text{g} \times \frac{1 \, \text{mol Mg}}{24.31 \, \text{g Mg}} = 0.009579 \, \text{mol Mg}
\]
The final value of moles:
\[
\text{Moles} = 0.009564 \, \text{mol Mg}
\]
**Section i: Calculating \(\Delta H\)**
The enthalpy change (\(\Delta H\)) is calculated using the heat energy released or absorbed (\(q_{\text{rxn}}\)) and the moles of reactant.
Given data:
- \(q_{\text{rxn}} = 0.453 \, \text{kJ}\)
- Moles of Mg calculated earlier = 0.009564 mol
Formula:
\[
\Delta H = \frac{q_{\text{rxn}}}{\text{moles}}
\]
Calculation for \(\Delta H\):
\[
\Delta H = \frac{-0.453 \, \text{kJ}}{0.009564 \, \text{mol Mg}} = -47.36 \, \text{kJ/mol}
\]
An additional provided value:
\[
\Delta H_{\text{rxn2}} = 47.4 \, \text{kJ/mol}
\]
Note: The calculation detail indicates a negative sign associated with \(\Delta H\), signifying an exothermic reaction.
**Conclusion:**
These calculations are critical in determining the enthalpy change associated with a chemical reaction where Mg acts as the limiting reagent. This understanding is vital for experiments requiring precise thermal measurements.
![**Calculating \( q_{\text{rxn}} \) and \( \Delta H \) for the 1 M HCl / Mg Ribbon Reaction**
The energy released during the reaction (\( -q_{\text{rxn}} \)) is absorbed by the solution and the calorimeter. This implies that the energy absorbed by the solution (\( q_{\text{sol}} \)) plus the energy absorbed by the calorimeter (\( q_{\text{cal}} \)) equals the energy released by the reaction. The equation is given by:
\[ -q_{\text{rxn}} = (q_{\text{sol}} + q_{\text{cal}}) \]
or
\[ -q_{\text{rxn}} = -[(m_{\text{sol}} \times C_{\text{sol}} \times \Delta T) + (C_{\text{cal}} \times \Delta T)] \]
**Note:** \( T_{\text{max}} \) is the final temperature for all species. For converting the volume of the solution to grams, use the density: density of 1 M HCl is 1.01 g/mL. The specific heat of the solution is 4.04 J/g°C.
**Calculate the following:**
a) **Volume of HCl (from data):**
\( V_{\text{HCl}} = 100.0 \, \text{mL} \)
b) **Find the mass of the solution using the mL of HCl and the density:**
\[ 100.0 \, \text{mL} \times 1.01 \, \text{g/mL} = 101 g \]
\( m_{\text{sol}} = 101.0 \, \text{g} \)
c) **Find the initial temperature of the HCl (from data):**
\[ T_{\text{initial}} = 20.2^\circ \text{C} \]
d) **Find the maximum reaction temperature (from data):**
\[ T_{\text{max}} = 33.2^\circ \text{C} \]
e) **Calculate \( \Delta T \):**
\[ \Delta T = (T_{\text{max}} - T_{\text{initial}}) \]
\[ \Delta T = 33.2^\circ \text{C} - 20.2^\circ \text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facf8e957-4fbe-471c-af8b-521f0436b36d%2F4d9d6182-74a4-4a00-81f8-fff6de475f6e%2Fmfwtinc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating \( q_{\text{rxn}} \) and \( \Delta H \) for the 1 M HCl / Mg Ribbon Reaction**
The energy released during the reaction (\( -q_{\text{rxn}} \)) is absorbed by the solution and the calorimeter. This implies that the energy absorbed by the solution (\( q_{\text{sol}} \)) plus the energy absorbed by the calorimeter (\( q_{\text{cal}} \)) equals the energy released by the reaction. The equation is given by:
\[ -q_{\text{rxn}} = (q_{\text{sol}} + q_{\text{cal}}) \]
or
\[ -q_{\text{rxn}} = -[(m_{\text{sol}} \times C_{\text{sol}} \times \Delta T) + (C_{\text{cal}} \times \Delta T)] \]
**Note:** \( T_{\text{max}} \) is the final temperature for all species. For converting the volume of the solution to grams, use the density: density of 1 M HCl is 1.01 g/mL. The specific heat of the solution is 4.04 J/g°C.
**Calculate the following:**
a) **Volume of HCl (from data):**
\( V_{\text{HCl}} = 100.0 \, \text{mL} \)
b) **Find the mass of the solution using the mL of HCl and the density:**
\[ 100.0 \, \text{mL} \times 1.01 \, \text{g/mL} = 101 g \]
\( m_{\text{sol}} = 101.0 \, \text{g} \)
c) **Find the initial temperature of the HCl (from data):**
\[ T_{\text{initial}} = 20.2^\circ \text{C} \]
d) **Find the maximum reaction temperature (from data):**
\[ T_{\text{max}} = 33.2^\circ \text{C} \]
e) **Calculate \( \Delta T \):**
\[ \Delta T = (T_{\text{max}} - T_{\text{initial}}) \]
\[ \Delta T = 33.2^\circ \text{C} - 20.2^\circ \text{
Expert Solution
Step 1
The heat released or absorbed (q) when the temperature of any substance is raised is calculated as:
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