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Chapter 6: Problem 109

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 \(\mathrm{M} \mathrm{KOH}\) ?

Short Answer

Expert verified
In conclusion, 4.20 kJ of energy will be released when 200.0 mL of 0.400 M HNO3 is mixed with 150.0 mL of 0.500 M KOH.

Step by step solution

01

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the neutralization reaction between HNO3 (a strong acid) and KOH (a strong base). The balanced chemical equation is: \(HNO_{3} + KOH \rightarrow H_{2}O + KNO_{3}\)
02

Calculate the moles of reactants

Next, we need to find the moles of HNO3 and KOH that are present in the given volumes and concentrations. Moles = Molarity × Volume (in Liters) For HNO3: Moles of HNO3 = 0.400 mol/L × 200.0 mL × (1 L / 1000 mL) = 0.0800 mol For KOH: Moles of KOH = 0.500 mol/L × 150.0 mL × (1 L / 1000 mL) = 0.0750 mol
03

Determine the limiting reactant

Now, we need to determine the limiting reactant by comparing the mole ratio of reactants to the coefficients in the balanced chemical equation. Mole ratio of HNO3 to KOH = moles of HNO3 / moles of KOH = 0.0800 mol / 0.0750 mol = 1.067 The balanced chemical equation shows a 1:1 mole ratio for the reactants. In this case, KOH has fewer moles than required for reacting with all HNO3 and thus, we identify KOH as the limiting reactant.
04

Calculate the moles of water produced

Once we have determined the limiting reactant, we can calculate the moles of water produced in the reaction. The balanced chemical equation shows a 1:1 mole ratio between reactants and water, so the moles of water produced will be equal to the moles of the limiting reactant. Moles of water produced = moles of the limiting reactant (KOH) Moles of water produced = 0.0750 mol
05

Calculate the energy released in the reaction

Finally, we can calculate the energy released in the reaction using the enthalpy of neutralization given, which is -56 kJ/mol of water produced. Energy released = moles of water produced × enthalpy of neutralization per mole of water produced Energy released = 0.0750 mol × (-56 kJ/mol) Energy released = -4.20 kJ In conclusion, 4.20 kJ of energy will be released when 200.0 mL of 0.400 M HNO3 is mixed with 150.0 mL of 0.500 M KOH.

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