You started off correctly by calculating the moles of each reactant.
#"Moles of HCl" = 0.050 color(red)(cancel(color(black)("L HCl"))) × "0.500 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.0250 mol HCl"#
#"Moles of NaOH" = 0.050 color(red)(cancel(color(black)("L NaOH"))) × "0.500 mol NaOHl"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.0250 mol NaOH"#
The chemical equation is
#"HCl + NaOH → NaCl" + "H"_2"O"#
You have 0.0250 mol #"HCl"# and 0.050 mol #"NaOH"#, so you have 0.0250 mol of reaction.
There are two heats involved in this process:
#"heat from reaction + heat to warm calorimeter = 0"#
#color(white)(mmmmm)q_1color(white)(mmm) +color(white)(mmmmm) q_2color(white)(mmmmmml) = 0#
#color(white)(mmmm)nΔ_text(r)Hcolor(white)(mm) + color(white)(mmmll)C_text(tot)ΔT color(white)(mmmmm)= 0#
In this problem,
#n color(white)(mll)= color(white)(l)"0.0250 mol"#
#Δ_text(r)H = "-56 720 J·mol"^"-1"#
#ΔTcolor(white)(ll) = color(white)(l)"3.2 °C"#
∴ #0.0250color(red)(cancel(color(black)("mol"))) × ("-56 720 J·"color(red)(cancel(color(black)("mol"^"-1")))) + C_text(tot)× "3.2 °C" = 0#
#"-1418 J"color(white)(l) + C_text(tot) × "3.2 °C" = 0#
#C_text(tot) ="1418 J"/"3.2 °C" = "440 J/°C"#
