**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 kJThus, the quantity of heat released in the reaction is approximately -147.654 kJ.

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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

Section: Chapter Questions

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