Question

ty ah Run 2 Run 1 Average initial temperature of HCl and A NaOH solutions (°C) 29.8 32.4 Maximum temperature of HCl-NAOH solution (°C) 34.5 30.5 Maximum temperature change for reaction (°C) 13 Total volume of HCl plus NaOH solutions (mL) Mass of final NaCl solution (g) Heat produced from HCl-NaOH reaction (J) Moles of H,O formed in reaction (mol) Heat expected from moles of H,0 formed (J) Heat absorbed by calorimeter (J) Heat capacity of calorimeter (J/oC) Average heat capacity of calorimeter (J/°C) Volume of HCl solution (mL) A Molarity of HCl solution (M) 2.054 2.054 Volume of NaOH solution (mL) S3 S3 Molarity of NAOH solution (M) 2083 Initial temperature of HCl solution (°C) 22 2 1 Initial temperature of NaOH solution (°C) 22 225 Temperatures of solution (°C) at 0.25-min (15-sec) intervals 0.25 24 32 0.50 30 20-S 30-5 0.75 34.5 32 33 33 33 32.9 32 32 32 32 31S 31 5 31 31 1.00 1.25 1.50 30-5 1.75 2.00 2.25 30-Y BO 2.50 2.75 30 30 30 3.00 3.25 3.50 3.75 29 5 4.00

Answer 1

If you consider the proposed neutralization reaction:

HCl + NaOH ----> NaCl + H₂O ;

The following volumes of HCl and NaOH were mixed to form a new solution:

volume of HCl= 50 mL of 2.054 M HCl solution

volume of NaOH= 53 mL of 2.083 M NaOH solution

Then, the temperature was measured twice, for the initial solution of HCl and NaOH (1 and 2):

Initial temperature of HCl solution (0 minutes): 22°C and 21 °C; getting an average initial temperature: 21.5 °C

Initial temperature of NaOH solution (0 minutes): 22.5 °C and 22 °C; getting an average initial temperature: 22.25 °C

Then, the temperature was measured twice for the resulting solution at 0.25-min intervals. From these values, you can determinate:

Maximum temperature of HCl-NaOH solution: (1) 30.5 °C and (2) 34.5 °C; getting an average (if you want): 32.5 °C

Maximum temperature change for reaction (you must subtract the highest temperature value with the lowest value): (1) 30.5 °C- 24°C=6.5 °C; and (2) 34.5 °C - 31°C=3.5 °C.

Total volume of HCl plus NaOH solutions (you must add the aggregate volumes of each initial solution)= 50 mL HCl solution + 53 mL NaOH solution=103 mL solution

Mass of final NaCl solution

First, you must get the moles of HCl and NaOH presents in solution, knowing the volumes used of each initial solution:

Moles of HCl= 50 mL solution x 2.054 mol HCl/1000 mL solution= 0.103 mol HCl

Moles of NaOH= 53 mL solution x 2.083 mol Na=H/1000 mL solution= 0.110 mol NaOH

Then, you can determine by stoichiometry that the HCl is the limiting reagent, because it reacted completely and there was still NaOH to react. Then, you can determinate the number of moles obtained for NaCl; knowing that 1 mol of HCl produces 1 mol of NaCl, 0.103 moles of HCl will produce 0.103 moles of NaCl.

And finally you can calculate the mass of NaCl solution knowing that its molecular mass is: 58.44 g/mol

Mass of NaCl= 0.103 mol NaCl x (58.44 g NaCl/1 mol NaCl) = 6.02 g NaCl

Heat produced from HCl-NaOH reaction (or formation of NaCl)

The heat can be determined by the following formula:

q= C Δt ; q is heat, C is the heat capacity of NaCl and Δt is the change in temperature.

The heat capacity of a substance (C) is defined as the amount of heat that is required to raise the temperature of a given amount of substance by one degree Celsius. That is:

C=m s; m is the mass of the substance and s the specific heat. The specific heat (s) is defined as the amount of heat that is required to increase a degree celsius the temperature of a gram of the substance, and is found in tables

specific heat (s) NaCl= 0.879 J/g °C. So (between 1 and 2 only ΔT varies):

(1) q=C Δt =(m s) Δt= (6.02 g NaCl)(0.879 J/g °C)(6.5 °C)=34.395 J

(2) q=C Δt =(m s) Δt= (6.02 g NaCl)(0.879 J/g °C)(3.5 °C)=18.521 J

Moles of H₂O formed in the reaction

You can determine it by stoichiometry; knowing that 1 mol of HCl produces 1 mol of H₂O, 0.103 moles of HCl will produce 0.103 moles of H₂O.

And finally you can calculate the mass of H₂O knowing that its molecular mass is: 18.015 g/mol

Mass of H₂O= 0.103 mol H₂O x (18.015 g H₂O/1 mol H₂O) = 1.856 g H₂O

Heat espected from moles of H₂O formed

specific heat (s) H₂O= 4.184 J/g °C. So,

(1) q=C Δt =(m s) Δt=(1.856 g H₂O) (4.184 J/g °C) (6.5 °C)=50.476 J

(2) q=C Δt =(m s) Δt=(1.856 g H₂O) (4.184 J/g °C) (3.5 °C)= 27.179 J