Question

A student mixes 35.2 mL of a 3.11 M sodium hydroxide solution with 34.9 mL of 2.95 M hydrochloric acid. The temperature of the mixture rises 18.2°C. The density of the resulting solution is 1.00 ml and J has a specific heat capacity of 4.184 J The heat capacity of the calorimeter is 3.86 E. Part 1: (a) Identify the limiting reagent for the reaction. HCI Part 2 out of 3 W try of (b) Calculate the heat of reaction (in J). x 10 Enter your answer in scientific notation. By star and assc ques Next part

calculate the heat of the reaction then find the enthalpy of neutralization in kJ/mol.

Answer 1

since, molanity of Ports 3.11 M, sodium hydroxide (NaOH) is 3.11 mole NaOH Contain 1ooo mL solution 35.2 mL 11 3.11 x 35.21 3.11 NaOH 0.1095 mole 1ooo NaOH 2.95 M hydrochloric hydrochloric acid (HCl) similarly, 34.9 mL 34.9 x 2.95 mole 1000 5 ) 0.1030 mole with mole neact Completely 0.1030 douard Hel will form mole out Nach v of 0.1030 0.10 95 mole) of Nach each o.1030 mole H20 and NaOH Hс 1 Nach +H₂O the As used up HO limiting reagent Completely, the reaction IHCl, Pant mass abonbed Heat heat of solution & temperature inctiease (b) by solution specific of solution density & Volume x Volume of solution) xl apecific heat) (increase Itemlematone) 35.2 +34.9) m2 x 1.00 g/mL me] 3/ a *c 5338.03 J x 4.184 goc x 18.2°c heat absorbed by Calorimelen - heat Capacity Calorimelen x increase temperatura (3-86 1 / ") (18.2 °C) x (18.2 °c) = 70.252 T

Hence, = heat abon bed total heat released the reaction by solution heat absorbed 70*252) J = J = 54o8.3 J by Calorimeten 5338.03 + 3 5.4083x10 3 heat of hence, 5.4083x10 ne action, 9 mon = Part 3 3 As 5.4083 xlo released J have been energy mole da oH for of by neutralizator O. 1030 01030 of mole enthally Ha, reaction, 5.4083 x 103 - 5.4083 Tlmol KIl mol 52.508 0.1030 0.1030 KJ/mol 52.508 Hence, enthalpy of neutralization KTlmol.