1. In a reaction, 90.00 mL of 0.90 M HC(a) and 110.00 mL of 1.00 M NaOH(aq) were
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Question
1. In a reaction, 90.00 mL of 0.90 M HC(a) and 110.00 mL of 1.00 M NaOH(aq) were added to a constant pressure calorimeter. The increase in temperature was 9.5 °C HCI(aq) + NaOH(aq) NaCl(aq) + HJX) AMin=???? a. Calculate the number of moles of HCl(aq) used. b. Calculate the number of moles of NaOH(a) used. c. Calculate the total volume of the solution (in mL). d. Assuming the density of the solution is 1.00 g/mL, calculate the mass of the solution, moln (in g) Assuming the specific heat of the solution is 4. 184 (in J). e. , calculate the heat absorbed by the solution, qsoln 8 "C Assuming the heat capacity of the calorimeter is 10.6 geal (in J). f. , calculate the heat absorbed by the calorimeter, g Calculate goa (in J) h. Calculate AHa (in kJ/mol) (Hint: divide gaa by moles of the limiting reactant). 2-1Explanation / Answer
(a)
Number of moles of HCl = 0.90 * 90.00 / 1000 = 0.081 mol
(b)
Number of moles of NaOH = 1.00 * 110.00 / 1000 = 0.11 mol
(c)
Total volume of the solution = 90.00 + 110.00 = 200.00 mL
(d)
Mass of solution = density * volume = 1.00 * 200.00 = 200.00 g.
(e)
Heat absorbed by solution = mass * specific heat * change in temperature
qsol = 200.00 * 4.184 * 9.5 = 7949.6 J
(f)
qcal = 10.6 * 9.5 = 100.7 J
(g)
qrxn = qsol + qcal = 7949.6 + 100.7 = 8050.3 J
(h)
According to the balanced equation,
HCl is limiting reagent.
deltaHrxn = qrxn / n = 8050.3 / 0.081 = 99386 J /mol
deltaHrxn = 99.4 kJ/mol
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