There are three heat transfers involved in this problem.
#"heat of neutralization + heat gained by solution" + "heat gained by calorimeter" = 0#
#q_1 + q_2 + q_3 = 0#
#nΔ_text(neut)H +mC_2ΔT + C_3ΔT= 0#
Let's calculate the heats separately.
#n = "0.135 mol"#
#q_1 = nΔ_text(neut)H = 0.135Δ_text(neut)Hcolor(white)(l) "mol"#
#m = "(50.0 + 50.0) g" = "100.0 g"#
#C_2 = "4.184 J·g·°C"^"-1"#
#ΔT = "(33.5 - 22.0) °C" = "11.5 °C"#
#q_2 = mC_2ΔT = 100.0 color(red)(cancel(color(black)("g"))) × "4.184J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 11.5 color(red)(cancel(color(black)("°C"))) = "4812 J"#
#q_3 = C_3ΔT = 20.6 "J"·color(red)(cancel(color(black)("°C"))) × 11.5 color(red)(cancel(color(black)("°C"))) = "236.9 J"#
#q_2 + q_3# represents the heat from the neutralization that was absorbed by the water and the calorimeter.
#q_2 + q_3 = "(4812 + 236.9) J = 5048 J"#
So, the total heat evolved is 5050 J (3 significant figures).
#q_1 = 0.135Δ_text(neut)Hcolor(white)(l) "mol" = -(q_2 + q_3) = "-5048 J"#
#Δ_text(neut)H = "-5048 J"/"0.135 mol" = "-37 400 J/mol" = "-37.4 kJ/mol"#
