es « previous | 5 of 19 next ome Exercise 7.26 The heat of neutralization of HCI
ID: 580584 • Letter: E
Question
es « previous | 5 of 19 next ome Exercise 7.26 The heat of neutralization of HCI(aq) by NaOH(ag) is 5584 kJ/mol H20 produced You may want to reference ( page) Section 7.1 while completing this problem Part A ents If 50.00 mL of 1.05 M NaOH is added to 25.00 mL of 1.84 M HCI, with both solutions originally at 24.66 °C, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02 g/mL and a specific heat of 3.98 Jg1C1) Tools Express your answer in degrees Celsius to four significant figures Area Settings T 33.25 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Provide FeedbackC Continue 21.11.2017 DELL F10 2 3 4 5Explanation / Answer
mol of HCl = MV = 1.84*(25*10^-3) = 0.046 mol
mol of NaOH = MV = 1.05*(50*10^-3) = 0.0525 mol
therefore, HCl is limiting reactant
Q = n*HRxN
Q = 0.046 *55.84 = 2.56864 kJ will be released
Qsolution = 2.56864 kJ = 2568.64 J
Q = m*C*(Tf-Ti)
mass = 50+25 = 75 mL
mass = D*V = 75*1.02 = 76.5 g
2568.64 = 76.5*3.98*(Tf-24.66)
Tf = 2568.64/( 76.5*3.98) + 24.66
Tf = 33.0964 °C
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