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\begin{tabular}{|l|l|} \hline Concentration of hydroiodic acid (M) & 2.12 \\ \hline Concentration of lithium hydroxide (M) & 2.05 \\ \hline Calorimeter constant (J/∘C) & 14.9 \\ \hline Volume of hydroiodic acid (mL) & 97.0 \\ \hline Temperature of hydroiodic acid (∘C) & 25.0 \\ \hline \end{tabular} \begin{tabular}{|l|l|} \hline Volume of lithium hydroxide (mL) & 91.5 \\ \hline Temperature of lithium hydroxide (∘C) & 25.0 \\ \hline Final Temperature after mixing (∘C) & 39.3 \\ \hline Mass of mixed solution (g) & 188.5 \\ \hline Moles of H+(mol) & 0.206 \\ \hline Moles of OH−(mol) & 0.188 \\ \hline Limiting Reactant & OH− \\ \hlineΔHncominiman ​(kJ/mol) & −57.5 \\ \hline \end{tabular}
Calculation 1: Heat absorbed/released by solution Geolution =mcΔT qsolution ​=(91.5 g+97.0 g)(4.184 J/g∘C)(39.3∘C−25.0∘C) qsolution ​=188.5 g(4.184 J/g∘C)(14.3∘C) qsolution ​=11278.18 J Calculation 2: Heat absorbed/released by calorimeter G calorimeter =cΔT qcalorimeter ​=14.9 J/∘C(39.3∘C−25.0∘C) q calsimeter =14.9 J/∘C(14.3∘C) qcalorimeter ​=213.07 J Calculation 3: Heat absorbed/released by reaction qrecleased ​+qabs sorbed ​=0 qreaction ​+qsolution ​+qcalorimetex ​=0 qreaction ​+11278.18 J+213.07 J=0 greaction =−11491.25 J Calculation 4: Moles of limiting reactant Calculation 5: Enthalpy of Neutralization Thermochemical equation for neutralization That means that you must show the chemical reaction taking place between the strong acid (HCl) and the strong base (NaOH) and identify its products. Additionally you must provide the DeltaH of neutralization you calculated in the previous calculation.
Discuss whether the reaction was endo-or exothermic and how this was determined. Discuss the magnitude of the calculated delta H of neutralization (i.e: discuss whether a little or a lot of energy was released or absorbed from the neutralization reaction). Discuss which of the two reagents was found to be the limiting reactant and how this was determined. Discuss which components were considered the 'surroundings' and which were the 'system' in this particular experiment and make sure to explain your reasoning. Discuss HOW the First Law of Thermodynamics applied to this experiment.]
3. Heat absorbedireleased by reaction The heat absorbed/released by the reaction is equal in magnitude but opposite in sign to the total amount of heat absorbed/released by the solution and the calorimeter. 9released ​+qahsorbed ​=0 qteaction ​+qsokatios ​+qcaloeimeter ​=0 qreartican +10888 J+205.02 J=0 qreactsin ​=−11093.02 J (unfounded) 4. Thermochemical equation for neutralization The enthalpy of neutralization for the reaction of a strong acid and strong base is the enthalpy associated with the reaction of a hydrogen ion with a hydroxide ion to produce a water molecule. H+(aq)+OH−(aq)→H2​O(l)ΔHtevitraliration ​ 5. Moles of limiting reactant The enthalpy of neutralization is calculated from the moles of the limiting reactant. motH+=95.9mLHCl×1000 mL1 L​×1LHCt2.01 molHCt​×1 molHCt1HolH+​ molH+=0.193 molH molOH=98.3 mLNaOH×1000 mL1 L​×1 LNaOH2.19 molNaOH​×1 molNaOH1 molOH​ molOH=0.215 molOH In this example, H+is the limiting reactant. 6. Enthalpy of neutralization The enthalpy of neutralization is calculated using the moles of the limiting reactant and the heat released by the reaction. Mractwa moles of limiting reactant ×ΔHmatnalask ​ ΔHneutralization ​= moles of limiting reactant  Quaram ​=0.193 mol−11093.02 J​ ΔHmaxushaki ​=−57476.8mnJ​×1 m0l1 kJ​=−57.5molkJ​