A reaction between magnesium oxide and nitric acid is conducted in a coffee cup calorimeter. Below is the provided data, along with the balanced chemical equation for the reaction:
(Hint: You need to determine the limiting reactant in order to calculate the moles)

MgO(s) + 2 HNO3(aq) --> Mg(NO3)2(aq) + H2O(l)

Mass of MgO used	0.510 g
Concentration of HNO3	1.10 M
Volume of HNO3 used	200.0 mL
Initial temperature of solution	21.00 °C
Final temperature of solution	23.49 °C

Density of HNO3: 1.02 g/mL
Specific heat of solution (s): 4.184 J/oC.g
Mass of solution = mass of solute + mass of solvent

Determine the value for ΔHrxn, in kJ/mol.ΔHrxn= 

To calculate the enthalpy change (\( \Delta H_{\text{rxn}} \)) for the reaction between magnesium oxide (MgO) and nitric acid (HNO₃), we need to follow these steps:

### Step 1: Determine the moles of \( \text{MgO} \) and \( \text{HNO}_3 \)

#### Moles of \( \text{MgO} \):

The molar mass of \( \text{MgO} \) is calculated as:
\[
\text{Molar mass of MgO} = 24.305 \, (\text{Mg}) + 16.00 \, (\text{O}) = 40.305 \, \text{g/mol}
\]
So the moles of MgO used are:
\[
\text{Moles of MgO} = \frac{0.510 \, \text{g}}{40.305 \, \text{g/mol}} \approx 0.01265 \, \text{mol}
\]

#### Moles of \( \text{HNO}_3 \):

Given that the concentration of \( \text{HNO}_3 \) is 1.10 M and the volume is 200.0 mL:
\[
\text{Moles of HNO}_3 = \text{Concentration} \times \text{Volume (in L)} = 1.10 \, \text{mol/L} \times 0.200 \, \text{L} = 0.220 \, \text{mol}
\]

### Step 2: Determine the limiting reactant

From the balanced chemical equation:
\[
\text{MgO} + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2\text{O}
\]
The molar ratio is 1:2. Therefore, we need 2 moles of \( \text{HNO}_3 \) for every 1 mole of \( \text{MgO} \).

- The required moles of \( \text{HNO}_3 \) for 0.01265 mol of \( \text{MgO} \):
\[
\text{Required moles of HNO}_3 = 2 \times 0.01265 = 0.0253 \, \text{mol}
\]
- Since we have 0.220 mol of \( \text{HNO}_3 \), which is more than enough to react with all the \( \text{MgO} \), \( \text{MgO} \) is the limiting reactant.

### Step 3: Calculate the heat (\( q \)) released in the reaction

The heat released or absorbed in the reaction can be calculated using the equation:
\[
q = m \cdot s \cdot \Delta T
\]
Where:
- \( m \) is the mass of the solution,
- \( s \) is the specific heat capacity of the solution,
- \( \Delta T \) is the change in temperature.

#### Mass of the solution:

To calculate the mass of the solution, we use the density of the \( \text{HNO}_3 \) solution (1.02 g/mL). The volume of the solution is 200.0 mL, so the mass of the solution is:
\[
\text{Mass of solution} = 1.02 \, \text{g/mL} \times 200.0 \, \text{mL} = 204.0 \, \text{g}
\]

#### Temperature change (\( \Delta T \)):

\[
\Delta T = 23.49^\circ \text{C} - 21.00^\circ \text{C} = 2.49^\circ \text{C}
\]

#### Heat released (\( q \)):

\[
q = 204.0 \, \text{g} \times 4.184 \, \text{J/g}^\circ\text{C} \times 2.49^\circ \text{C} = 2128.4 \, \text{J} = 2.1284 \, \text{kJ}
\]

### Step 4: Calculate \( \Delta H_{\text{rxn}} \)

Since \( \text{MgO} \) is the limiting reactant, the heat calculated is for the amount of \( \text{MgO} \) that reacted (0.01265 mol). The enthalpy change per mole is:
\[
\Delta H_{\text{rxn}} = \frac{q}{\text{moles of MgO}} = \frac{2.1284 \, \text{kJ}}{0.01265 \, \text{mol}} \approx 168.3 \, \text{kJ/mol}
\]

Thus, the value of \( \Delta H_{\text{rxn}} \) is **-168.3 kJ/mol** (the negative sign indicates that the reaction is exothermic).

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Let me know if you need further clarification or a breakdown of any steps!

### Relative Questions:
1. How do you determine the limiting reactant in any chemical reaction?
2. What is the importance of using a calorimeter in enthalpy calculations?
3. Why is it necessary to convert mL to L when calculating moles for solutions?
4. How does the density of a solution affect the calculation of heat in calorimetry?
5. How would the result change if the final temperature was higher?

### Tip:
Always ensure that units are consistent throughout your calculations to avoid errors in the final result.

A reaction between magnesium oxide and nitric acid is conducted in a coffee cup calorimeter. Below is the provided data, along with the balanced chemical equation for the reaction: (Hint: You need to determine the limiting reactant in order to calculate the moles) MgO(s) + 2 HNO3(aq) --> Mg(NO3)2(aq) + H2O(l) Mass of MgO used 0.510 g Concentration of HNO3 1.10 M Volume of HNO3 used 200.0 mL Initial temperature of solution 21.00 °C Final temperature of solution 23.49 °C Density of HNO3: 1.02 g/mL Specific heat of solution (s): 4.184 J/oC.g Mass of solution = mass of solute + mass of solvent Determine the value for ΔHrxn, in kJ/mol.

Learn how to calculate the enthalpy change (ΔHrxn) for the reaction between magnesium oxide and nitric acid in a coffee cup calorimeter. This problem involves key concepts like stoichiometry, limiting reactants, and heat calculation using the specific heat capacity formula. Suitable for Grades 11-12.

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