2. In an acid-base neutralization reaction, 50.00 mL of 1.00 M HCI was mixed wit

Question

2. In an acid-base neutralization reaction, 50.00 mL of 1.00 M HCI was mixed with 50.00 ml of 1.00 M NaOH. Calculate the number of moles of HCl(aq) and NaOH(aq) in the above samples. Also calculate the number of moles of each of the products formed by the above reaction. Is there a limiting reagent for the above reaction? If so, which one? Which one is the excess reagent? Show all work. a. b. Using ??,"values from standard tables (cite your source), calculate the enthalpy of neutralization for the reaction between 1.0 mole of HCl and 1.0 mole of NaOH. Use this information to calculate the enthalpy of neutralization for the reaction in 2a above. Show all work. c. Thermo

Explanation / Answer

Number of moles of HCl = Volume of solution (in L) * Molarity = 50/1000L * 1M = 0.05 moles

Number of moles of NaOH = Volume of solution (in L) * Molarity = 50/1000L * 1M = 0.05 moles

The reaction happening will be

HCl(aq) + NaOH(aq) ---------- NaCl(aq) + H2O

Number of moles of NaCl formed = Number of moles of H2O formed = Number of moles of NaOH = Number of moles of HCl = 0.05 moles

a) There is no limiting reagent for the above reaction, since number of moles of HCl and number of moles of NaOH are same

Since both the reactants will be fully used by the end of the experiment, there will be no moles left of the reactants

Hence there is no limiting reagent and no excess reagent

b) Enthalpy of neutralization = -57.9 kJ/mol

(Data taken from my course book)

c) Enthalpy of neutralization of above reaction = number of moles * molar enthalpy of neutralization

=> 0.05 mol * (-57.9 kJ/mol)

=> -2.895 kJ

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