Explanation:
The heat gained by the solution = q
![q=mc\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=q%3Dmc%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
where,
q = heat gained = ?
c = specific heat of solution= ![4.18 J/^oC](https://tex.z-dn.net/?f=4.18%20J%2F%5EoC)
Mass of the solution(m) = mass of water + mass of calcium chloride
Mass of water = ?
Volume of water = 50.00 mL
Density of water = 1.00 g/mL
Mass = Density × Volume
m = 1.00 g/mL × 50.00 mL = 50.00 g
Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g
= final temperature = ![25.8 ^oC](https://tex.z-dn.net/?f=25.8%20%5EoC)
= initial temperature = ![22.6 ^oC](https://tex.z-dn.net/?f=22.6%20%5EoC)
Now put all the given values in the above formula, we get:
![q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC](https://tex.z-dn.net/?f=q%3D54.51%20g%5Ctimes%204.18%20J%2Fg%5EoC%5Ctimes%20%2825.8-22.6%29%5EoC)
![q=729.126 J](https://tex.z-dn.net/?f=q%3D729.126%20J)
The heat gained by the solution is 729.126 J.
Heat energy released during the reaction = q'
q' = -q ( law of conservation of energy)
q' = -729.126 J
The heat energy released during the reaction is -729.126 J.
Moles of calcium chloride, n = ![\frac{4.51 g}{111 g/mol}=0.04063 mol](https://tex.z-dn.net/?f=%5Cfrac%7B4.51%20g%7D%7B111%20g%2Fmol%7D%3D0.04063%20mol)
![\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Cfrac%7Bq%27%7D%7Bn%7D%3D%5Cfrac%7B-729.126%20J%7D%7B0.04063%20mol%7D%3D-17%2C945.23%20J%2Fmol%3D%20-17.945%20kJ%2Fmol)
The ΔH of the reaction is -17.945 kJ/mol.