Question

Calculate the heat loss or heat gain of the 3 solution mixtures (qrxn). Reactions are: NaOH + HCl ---> H2O+NaCl NaOH+NH4Cl--> NH3+NaCl+H2O NH3 + HCl --> NH4Cl

Answer 1

lets write the reactions clearly first:

NaOH + HCl ---> H2O+NaCl

NaOH+NH4Cl--> NH3+NaCl+H2O

NH3 + HCl --> NH4Cl

adding all 3 equations   

2 NaOH + 2HCl ----->2H₂O + 2NaCl

this is the final expression we got after adding all the reactions and heat of reaction of this reaction will give heat gain or loss

heat of reaction = enthalpy of formation of products - enthalpy of formation of reactants

= 2 mol x −241.818 kJ/mol + (−407.27kJ/mol) x 2 mol -{−167.2 kJ/mol x 2mol + (−469.15kJ/mol x 2 mol)}

= -25.476 kJ