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.