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Chapter 6: Problem 103

The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{~kJ} / \mathrm{mol}\) water produced. How much energy will be released when \(200.0 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(150.0 \mathrm{~mL}\) of \(0.500 M \mathrm{KOH}\) ?

Short Answer

Expert verified
The amount of energy released when $200.0 \mathrm{~mL}$ of $0.400 \mathrm{M} \mathrm{HNO}_{3}$ is mixed with $150.0 \mathrm{~mL}$ of $0.500 \mathrm{M} \mathrm{KOH}$ is $-4.2 \mathrm{~kJ}$.

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction of nitric acid (HNO3) with potassium hydroxide (KOH) is: \( HNO_3 + KOH \rightarrow KNO_3 + H_2O \)
02

Calculate the number of moles of HNO3 and KOH

Given the volume and molar concentration of HNO3 and KOH, we can calculate the number of moles of each reactant: Moles of HNO3 = (volume of HNO3) × (molar concentration of HNO3) = (200.0 mL) × (0.400 mol/L) = 0.200 L × 0.400 mol/L = 0.080 mol Moles of KOH = (volume of KOH) × (molar concentration of KOH) = (150.0 mL) × (0.500 mol/L) = 0.150 L × 0.500 mol/L = 0.075 mol
03

Determine the limiting reactant and calculate the moles of water produced

Now, we will compare the mole ratio of HNO3 and KOH to determine the limiting reactant, and use that to calculate the moles of water produced. Mole ratio of HNO3 to KOH = (moles of HNO3) / (moles of KOH) = 0.080 mol / 0.075 mol = 1.067 Since the mole ratio of HNO3 to KOH in the balanced chemical equation is 1:1, we can see that KOH is the limiting reactant. Therefore, the moles of water produced will equal the moles of KOH: Moles of water produced = moles of KOH = 0.075 mol
04

Calculate the energy released using the enthalpy of neutralization

We are given the enthalpy of neutralization per mole of water produced, which is -56 kJ/mol. To calculate the energy released, we will multiply this value by the moles of water produced: Energy released = (enthalpy of neutralization) × (moles of water produced) = (-56 kJ/mol) × (0.075 mol) = -4.2 kJ Thus, the amount of energy released when 200.0 mL of 0.400 M HNO3 is mixed with 150.0 mL of 0.500 M KOH is -4.2 kJ.

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