There are two heat transfers involved in this problem.
#"heat of neutralization + heat gained by solution" = 0#
#color(white)(mmmmm)q_1color(white)(mmmmll) + color(white)(mmmmmm)q_2 color(white)(mmmml)= 0#
#color(white)(mmml)nΔ_text(neut)Hcolor(white)(mmm) +color(white)(mmmmll) mCΔTcolor(white)(mmm)= 0#
(a) Heat of neutralization
The net ionic equation for the reaction is
#"H"_3"O"^"+" + "OH"^"-" → 2"H"_2"O"#
#"Moles of H"_3"O"^"+" = "0.020 60" color(red)(cancel(color(black)("L H"_2"SO"_4))) × (0.5000 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) × ("2 mol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) ="0.020 60 mol H"_3"O"^"+"#
#"Moles of OH"^"-" = "0.020 60" color(red)(cancel(color(black)("L NaOH"))) × (1.000 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × ("1 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.020 60 mol OH"^"-"#
You have equal moles of #"H"_3"O"^"+"# and of #"OH"^"-"#, so you have #"0.020 60 mol reaction"#.
∴ #n = "0.020 60 mol"#
#q_1 = nΔ_text(neut)H = "0.020 60"Δ_text(neut)Hcolor(white)(l) "mol"#
(b) Heat gained by water
#m = "(20.60 + 20.60) g" = "41.20 g"#
#C = "4.184 J·g·°C"^"-1"#
#ΔT = T_2 - T_1 = "(30.17 - 23.50) °C" = "6.67 °C"#
#q_2 = mCΔT = 41.20 color(red)(cancel(color(black)("g"))) × "4.184J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 6.67 color(red)(cancel(color(black)("°C"))) = "1150 J"#
(c) Molar heat of neutralization
#nΔ_text(neut)H + mCΔT = "0.020 60"Δ_text(neut)Hcolor(white)(l) "mol" + "1150 J" = 0#
#Δ_text(neut)H = "-1150 J"/"0.020 60 mol" = "-55 800 J/mol" = "-55.8 kJ/mol"#
