Question

inMACII Page 1: Question 4 (1 point) 2 3 1 The aluminum cup inside your calorimeter weighs 41.55 g. You add 59.21 g of 1.0 M acetic acid solution and 50.03 g of 1.0 M sodium hydroxide solution to the calorimeter. Both solutions have an initial temperature of 19.9 °C, and the final 5 4 temperature after addition is 26.8 °C. What is the molar enthalpy of neutralization, in units of kJ/mol? Assume that: the calorimeter is completely insulated the heat capacity of the empty calorimeter is the heat capacity of the aluminum cup: 0.903 J g1 °C-1 the density of the two solutions is the same as that of water: 1.00 g/mL. the heat capacity of the two solutions is the same as that of water: 4.184 J g1 °C-1 Perform all calculations without rounding, but then provide your answer to the correct number of significant figures. Your Answer: units Answer

Answer 1

sheah (0) = M COT mass = 59.21 g + 50.03 g = 109.24g Specitic heah (C) = 4,184 3/g.c. AT= 26.8°C - 19.9°C = 6.9°C 0-109.24 g x 4. 1845/9.1 x 6.9°C - 3 153.1153 The enthalpy change at for the reaction is 3 153.7155 STEP amalar enthal by of neutralization equation for the reaction is CH3COOH(aq) + NaOH(aq) - CH₃COONO faq) + H2O(l) males of CH₂ (OH- malority & volume Volume = 59.214 1.00g - 590im isloog - 59.21 m2 = 0.059214 narea mole 0.059211 x 0. 1 mol = 0.6592) mar male of rooh is the rate limiting reagent since it has relafinely few number of moles To calculate the molar enthalpy of neutralization we afhentralization - heah Change Number of male - 3153.7155 - 63036.478 3 0.05003 mole but 1ooo Talk so - 63.036 A3 - mod correct significant figa 63.04 Alone may