What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.631 mol of CO₂ in the following chemical reaction? SrO (s) + CO₂ (g) → SrCO3 (s) AH° = -234 kJ/ mol
What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.631 mol of CO₂ in the following chemical reaction? SrO (s) + CO₂ (g) → SrCO3 (s) AH° = -234 kJ/ mol
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
Chapter1: Chemical Foundations
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![**Question:**
What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.631 mol of CO₂ in the following chemical reaction?
\[ \text{SrO (s) + CO}_2 \text{ (g)} \rightarrow \text{SrCO}_3 \text{ (s)} \quad \Delta H^\circ = -234 \, \text{kJ/mol} \]
**Explanation:**
This question is about calculating the quantity of heat released during a chemical reaction between strontium oxide (SrO) and carbon dioxide (CO₂) to form strontium carbonate (SrCO₃). The reaction is exothermic, and each mole of SrO reacting releases -234 kJ of heat.
**Given:**
- Amount of SrO: 1.20 mol
- Amount of CO₂: 0.631 mol
- Enthalpy change (ΔH°): -234 kJ/mol
**Calculation Steps:**
1. **Determine the Limiting Reactant:**
- The stoichiometry of the reaction is 1:1 for SrO and CO₂.
- Since 0.631 mol of CO₂ is less than the 1.20 mol of SrO, CO₂ is the limiting reactant.
2. **Calculate the Heat Released:**
- Heat released = moles of limiting reactant × ΔH°
- Heat released = 0.631 mol × (-234 kJ/mol)
- Heat released = -147.654 kJ
Thus, the quantity of heat released in the reaction is approximately -147.654 kJ.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff51b0c15-3061-49f7-95de-f69fd3531e7d%2F627f360b-51fb-42ba-a73e-33a557514415%2F5c9i7ru_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.631 mol of CO₂ in the following chemical reaction?
\[ \text{SrO (s) + CO}_2 \text{ (g)} \rightarrow \text{SrCO}_3 \text{ (s)} \quad \Delta H^\circ = -234 \, \text{kJ/mol} \]
**Explanation:**
This question is about calculating the quantity of heat released during a chemical reaction between strontium oxide (SrO) and carbon dioxide (CO₂) to form strontium carbonate (SrCO₃). The reaction is exothermic, and each mole of SrO reacting releases -234 kJ of heat.
**Given:**
- Amount of SrO: 1.20 mol
- Amount of CO₂: 0.631 mol
- Enthalpy change (ΔH°): -234 kJ/mol
**Calculation Steps:**
1. **Determine the Limiting Reactant:**
- The stoichiometry of the reaction is 1:1 for SrO and CO₂.
- Since 0.631 mol of CO₂ is less than the 1.20 mol of SrO, CO₂ is the limiting reactant.
2. **Calculate the Heat Released:**
- Heat released = moles of limiting reactant × ΔH°
- Heat released = 0.631 mol × (-234 kJ/mol)
- Heat released = -147.654 kJ
Thus, the quantity of heat released in the reaction is approximately -147.654 kJ.
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