What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.951 mol of CO₂ in the Following chemical reaction?

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Chemical Reaction Energy Calculation**

**Problem Statement:**

What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.951 mol of CO₂ in the following chemical reaction?

\[ \text{SrO (s) + CO}_{2}\text{ (g) } \rightarrow \text{SrCO}_{3}\text{ (s)} \]

**Reaction Enthalpy:** 

\[ \Delta H^\circ = -234 \, \text{kJ/mol} \]

**Solution Overview:**

To find the quantity of heat released:

1. Identify the limiting reactant by comparing the mole ratio of the reactants to the stoichiometry of the reaction.

2. Calculate the heat released using the limiting reactant and the given enthalpy change (\( \Delta H^\circ \)).

**Detailed Steps:**

- Stoichiometrically, 1 mole of SrO reacts with 1 mole of CO₂ to form 1 mole of SrCO₃.
- Given 1.20 mol of SrO and 0.951 mol of CO₂, CO₂ is the limiting reactant.
- Using the enthalpy change, the heat released is calculated based on the moles of the limiting reactant:

\[ \text{Heat released} = 0.951 \, \text{mol} \times (-234 \, \text{kJ/mol}) = -222.834 \, \text{kJ} \]

**Conclusion:**

The quantity of heat released in this reaction is 222.834 kJ.
Transcribed Image Text:**Chemical Reaction Energy Calculation** **Problem Statement:** What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.951 mol of CO₂ in the following chemical reaction? \[ \text{SrO (s) + CO}_{2}\text{ (g) } \rightarrow \text{SrCO}_{3}\text{ (s)} \] **Reaction Enthalpy:** \[ \Delta H^\circ = -234 \, \text{kJ/mol} \] **Solution Overview:** To find the quantity of heat released: 1. Identify the limiting reactant by comparing the mole ratio of the reactants to the stoichiometry of the reaction. 2. Calculate the heat released using the limiting reactant and the given enthalpy change (\( \Delta H^\circ \)). **Detailed Steps:** - Stoichiometrically, 1 mole of SrO reacts with 1 mole of CO₂ to form 1 mole of SrCO₃. - Given 1.20 mol of SrO and 0.951 mol of CO₂, CO₂ is the limiting reactant. - Using the enthalpy change, the heat released is calculated based on the moles of the limiting reactant: \[ \text{Heat released} = 0.951 \, \text{mol} \times (-234 \, \text{kJ/mol}) = -222.834 \, \text{kJ} \] **Conclusion:** The quantity of heat released in this reaction is 222.834 kJ.
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