o tecuibied Raeaiaa firal wth 197 0 Enthalpies of Reaction 5.4: Enthalpy of Neut
ID: 553909 • Letter: O
Question
o tecuibied Raeaiaa firal wth 197 0 Enthalpies of Reaction 5.4: Enthalpy of Neutralization 4,164 Balanced equation for the reaction Identity of the acid Molarity of the acid Volume of the acid Moles of the acid Identification of the base Molarity of the base Volume of the base Moles of the base 25ml 20 25 mL 21 T. 22 Average T for the acid and base Mass of the solution (Mass" er,' Density-m" Volutnenanx ) Density of the reaction mixture is 1.00 g/mL 25 26 27 Moles osiin 29 Moles0 ond As always, attach your sample calculations. Calculations were performed for the shaded rowsExplanation / Answer
The equation for the reaction is
HCl (aq)+ NaOH(aq) ------> NaCl + HO(l)
20) Moles of acid = Molarity of acid x volume of acid in L
= 1.0 mol/L x (25/1000)L = 0.0250
21) Moles of base = Molarity of base x volume of base in L
= 1.10 mol/L x (25/1000)L = 0.0275
22) Average Tinitial for acid and base = (Tinitial for acid + Tinitial for base)/2
= (23.3 + 21)/2 °C = 22.15°C
23) T=Tfinal – Tinitial = (26.2 – 22.15) °C = 4.05 °C
24) Volume of solution = (25.0 + 25.0) mL = 50.0 mL
Mass of solution = Volume x density
= 50.0 ml × 1.0 g/ml = 50 g
25) q cal = cannot be calculated, because the heat capacity of the calorimeter is not given.
26) qsolution = mcT
= (50 g × 4.184 J·g¹°C¹ × 4.05 °C)
= 847.26 J
27) q rxn = q cal is not given, so it can’t be calculated.
Moles of limiting reagent = acid is limiting reagent. Its mol = 0.0250
Moles of H2O formed = 0.0250
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