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"
