100. mL sample of 0.200M aq HCl acid is added to 100mLof 0.200M aq ammonia in a calorimeter.whose heat capacity is unknown. The following reaction occurs when the 2 solutions are mixed HCl (aq)+NH3 ---> NH4Cl(aq) The temperature increase is 5°C calculate (delta)H per mole of HCl and NH3 reacted

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A 100. mL sample of 0.200M aq HCl acid is added to 100mLof 0.200M aq ammonia in a calorimeter.whose heat capacity is unknown. The following reaction occurs when the 2 solutions are mixed HCl (aq)+NH3 ---> NH4Cl(aq) The temperature increase is 5°C calculate (delta)H per mole of HCl and NH3 reacted
**Title: Calculating Enthalpy Change for a Neutralization Reaction**

**Experiment Overview:**

In this experiment, a 100 mL sample of 0.200 M aqueous hydrochloric acid (HCl) is combined with a 100 mL sample of 0.200 M aqueous ammonia (NH₃) in a calorimeter. The calorimeter's heat capacity is unspecified.

**Chemical Reaction:**

\[
\text{HCl(aq)} + \text{NH}_3(\text{aq}) \rightarrow \text{NH}_4\text{Cl(aq)}
\]

**Observations:**

Upon mixing these two solutions, the temperature of the mixture rises by 2.34°C.

**Objective:**

Calculate the enthalpy change (\(\Delta H\)) per mole of HCl and NH₃ reacted.

**Calculation Steps:**

1. **Formula Used:**
   \[
   q = m \times s \times \Delta T
   \]
   Where:
   - \(q\) is the heat absorbed or released,
   - \(m\) is the mass of the solution (assuming density similar to water, therefore approximately 200 g for 200 mL),
   - \(s\) is the specific heat capacity (assuming it's similar to water, \(4.18 \, \text{J/g°C}\)),
   - \(\Delta T\) is the change in temperature (2.34°C).

2. **Determine \(\Delta H\):**
   The enthalpy change per mole can be calculated using the formula:
   \[
   \Delta H = \frac{q}{n}
   \]
   Where:
   - \(n\) is the number of moles of the limiting reactant (found by multiplying the concentration by the volume for one of the solutions).

**Additional Notes:**

- The calculation should take into account whether there is any heat absorbed by the calorimeter itself if specified.
- Ensure that volumes and concentrations are in compatible units for accurate calculations.

This experiment demonstrates an exothermic neutralization reaction where heat is released, leading to an increase in temperature, which can be quantified to find the enthalpy change.
Transcribed Image Text:**Title: Calculating Enthalpy Change for a Neutralization Reaction** **Experiment Overview:** In this experiment, a 100 mL sample of 0.200 M aqueous hydrochloric acid (HCl) is combined with a 100 mL sample of 0.200 M aqueous ammonia (NH₃) in a calorimeter. The calorimeter's heat capacity is unspecified. **Chemical Reaction:** \[ \text{HCl(aq)} + \text{NH}_3(\text{aq}) \rightarrow \text{NH}_4\text{Cl(aq)} \] **Observations:** Upon mixing these two solutions, the temperature of the mixture rises by 2.34°C. **Objective:** Calculate the enthalpy change (\(\Delta H\)) per mole of HCl and NH₃ reacted. **Calculation Steps:** 1. **Formula Used:** \[ q = m \times s \times \Delta T \] Where: - \(q\) is the heat absorbed or released, - \(m\) is the mass of the solution (assuming density similar to water, therefore approximately 200 g for 200 mL), - \(s\) is the specific heat capacity (assuming it's similar to water, \(4.18 \, \text{J/g°C}\)), - \(\Delta T\) is the change in temperature (2.34°C). 2. **Determine \(\Delta H\):** The enthalpy change per mole can be calculated using the formula: \[ \Delta H = \frac{q}{n} \] Where: - \(n\) is the number of moles of the limiting reactant (found by multiplying the concentration by the volume for one of the solutions). **Additional Notes:** - The calculation should take into account whether there is any heat absorbed by the calorimeter itself if specified. - Ensure that volumes and concentrations are in compatible units for accurate calculations. This experiment demonstrates an exothermic neutralization reaction where heat is released, leading to an increase in temperature, which can be quantified to find the enthalpy change.
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