Question

actual mass of kcl used
moles of kcl used
mass of water
initial temperature, ( t_i, ^ { circ } c )
final temperature, ( t_f, ^ { circ } c )
(min. or max. from data above)
( delta t, ^ { circ } c )
( q _ { \text { system } }, j )
( delta h, kj / mol )
mean ( delta h, kj / mol )

Explanation:

Step1: Calculate ΔT

ΔT = Tf - Ti
For Trial 1: ΔT = 20.9 - 21.9 = - 1.0°C
For Trial 2: ΔT = 20.8 - 21.8 = - 1.0°C

Step2: Calculate qsystem

q = m c ΔT, where c = 4.18 J/g°C, m = mass of solution (mass of KCl + mass of water)
For Trial 1: m = 1.860 + 100.44 = 102.3 g
q1 = 102.3 4.18 (-1.0) = - 428 J
For Trial 2: m = 1.849 + 100.78 = 102.6 g
q2 = 102.6 4.18 (-1.0) = - 429 J

Step3: Calculate ΔH

ΔH = q / n, where n is moles of KCl
For Trial 1: ΔH1 = - 428 / 0.025 = - 17120 J/mol = - 17.1 kJ/mol
For Trial 2: ΔH2 = - 429 / 0.025 = - 17160 J/mol = - 17.2 kJ/mol

Step4: Calculate Mean ΔH

Mean ΔH = (ΔH1 + ΔH2) / 2 = (- 17.1 - 17.2) / 2 = - 17.15 kJ/mol

Answer:

Trial 1 ΔT: - 1.0°C, qsystem: - 428 J, ΔH: - 17.1 kJ/mol
Trial 2 ΔT: - 1.0°C, qsystem: - 429 J, ΔH: - 17.2 kJ/mol
Mean ΔH: - 17.15 kJ/mol