Respuesta :
Explanation:
The heat gained by the solution = q
[tex]q=mc\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = ?
c = specific heat of solution= [tex]4.18 J/^oC[/tex]
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
[tex]T_{final}[/tex] = final temperature = [tex]25.8 ^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]22.6 ^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC[/tex]
[tex]q=729.126 J[/tex]
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 = [tex]\frac{4.51 g}{111 g/mol}=0.04063 mol[/tex]
[tex]\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol[/tex]
The ΔH of the reaction is -17.945 kJ/mol.
