Calculate the heat of reaction (in J). Round your answer to 3 significant digits. reaction - J X Ś

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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need help with calculating heat of reaction

**Educational Website Transcription:**

**Title: Calculating the Heat of Reaction**

**Section: Identifying the Limiting Reagent**

A critical step in determining the heat of reaction involves identifying the limiting reagent. In the current example, sodium hydroxide (NaOH) has been identified as the limiting reagent for the reaction.

**Progress Indicator:**
- Current Task: Part 1 of 4
- Progress Bar: Visible indication of task completion status.

**Section: Calculating the Heat of Reaction**

*Task: Calculate the heat of reaction (in Joules). Round your answer to three significant digits.*

- **Formula Displayed:** The heat of reaction is denoted by \( q_{\text{reaction}} \) and is measured in Joules (J).
- **Input Field:** A text box is provided for entering the numerical value of the calculated heat of reaction. The format specifies multiplying by a power of ten, indicated by a box marked \(\times 10^{}\).

**Navigation:**
- A "Next Part" button is available for proceeding to the subsequent task.

**Note:** No graph or diagram is present in this section.
Transcribed Image Text:**Educational Website Transcription:** **Title: Calculating the Heat of Reaction** **Section: Identifying the Limiting Reagent** A critical step in determining the heat of reaction involves identifying the limiting reagent. In the current example, sodium hydroxide (NaOH) has been identified as the limiting reagent for the reaction. **Progress Indicator:** - Current Task: Part 1 of 4 - Progress Bar: Visible indication of task completion status. **Section: Calculating the Heat of Reaction** *Task: Calculate the heat of reaction (in Joules). Round your answer to three significant digits.* - **Formula Displayed:** The heat of reaction is denoted by \( q_{\text{reaction}} \) and is measured in Joules (J). - **Input Field:** A text box is provided for entering the numerical value of the calculated heat of reaction. The format specifies multiplying by a power of ten, indicated by a box marked \(\times 10^{}\). **Navigation:** - A "Next Part" button is available for proceeding to the subsequent task. **Note:** No graph or diagram is present in this section.
Below is the transcribed data and explanation that might appear on an educational website:

---

### Chemistry Experiment: Reaction Between Sodium Hydroxide and a Monoprotic Acid

**Collected Data:**

- **Concentration of HA ([HA] M):** 4.75 M
- **Initial Temperature of HA (°C):** 21.0 °C
- **Volume of HA (mL):** 56.2 mL
- **Concentration of NaOH ([NaOH] M):** 2.45 M
- **Initial Temperature of NaOH (°C):** 21.0 °C
- **Volume of NaOH (mL):** 85.0 mL
- **Final Solution Temperature (°C):** 34.3 °C
- **Density of Solution (g/mL):** 1.70 g/mL
- **Specific Heat of Solution (J/g·°C):** 4.004 J/g·°C
- **Calorimeter Constant (J/°C):** 38.76 J/°C

**Task:**

1. **Identify the Limiting Reagent:** For this experiment, you need to determine the limiting reagent. In the provided data, the dropdown suggests that NaOH is the limiting reagent.

**Progress Indicator:**

- The interface displays progress through a "Part: 1 / 4" progress bar, indicating the experiment or task is part one of a four-part series.

**Instructions:**

- Analyze the data and determine the limiting reagent by using stoichiometric calculations and considering the initial concentrations and volumes of the reactants.

### Explanation of the Data:

The table provides initial experimental setup data and measured outcomes for the reaction between a monoprotic acid (HA) and Sodium Hydroxide (NaOH). The experiment begins with both reactants at the same initial temperature. A noticeable increase in the solution's final temperature reflects the exothermic nature of the reaction. Understanding the limiting reagent is crucial for predicting the extent of the reaction.

--- 

The educational platform uses this data to guide students through the process of identifying the limiting reagent, enhancing comprehension of chemical reactions and stoichiometry.
Transcribed Image Text:Below is the transcribed data and explanation that might appear on an educational website: --- ### Chemistry Experiment: Reaction Between Sodium Hydroxide and a Monoprotic Acid **Collected Data:** - **Concentration of HA ([HA] M):** 4.75 M - **Initial Temperature of HA (°C):** 21.0 °C - **Volume of HA (mL):** 56.2 mL - **Concentration of NaOH ([NaOH] M):** 2.45 M - **Initial Temperature of NaOH (°C):** 21.0 °C - **Volume of NaOH (mL):** 85.0 mL - **Final Solution Temperature (°C):** 34.3 °C - **Density of Solution (g/mL):** 1.70 g/mL - **Specific Heat of Solution (J/g·°C):** 4.004 J/g·°C - **Calorimeter Constant (J/°C):** 38.76 J/°C **Task:** 1. **Identify the Limiting Reagent:** For this experiment, you need to determine the limiting reagent. In the provided data, the dropdown suggests that NaOH is the limiting reagent. **Progress Indicator:** - The interface displays progress through a "Part: 1 / 4" progress bar, indicating the experiment or task is part one of a four-part series. **Instructions:** - Analyze the data and determine the limiting reagent by using stoichiometric calculations and considering the initial concentrations and volumes of the reactants. ### Explanation of the Data: The table provides initial experimental setup data and measured outcomes for the reaction between a monoprotic acid (HA) and Sodium Hydroxide (NaOH). The experiment begins with both reactants at the same initial temperature. A noticeable increase in the solution's final temperature reflects the exothermic nature of the reaction. Understanding the limiting reagent is crucial for predicting the extent of the reaction. --- The educational platform uses this data to guide students through the process of identifying the limiting reagent, enhancing comprehension of chemical reactions and stoichiometry.
Expert Solution
Step 1

Given that

Initial temperature = 21.0 °C

Final Temperature = 34.3 °C

Change in temperature = 34.3 °C – 21.0 = 13.3 °C

Total volume of the solution = 56.2 + 85.0 = 141.2 mL

Density of the solution = 1.70 g/mL

Mass of the solution = 141.2 mL * 1.70 g/mL = 240.04 g

Specific heat of the solution = 4.004 J/(g.°C)

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