# Question d8b91

Mar 29, 2017

The heat capacity of the system is 440 J/°C.

#### Explanation:

You started off correctly by calculating the moles of each reactant.

$\text{Moles of HCl" = 0.050 color(red)(cancel(color(black)("L HCl"))) × "0.500 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.0250 mol HCl}$

$\text{Moles of NaOH" = 0.050 color(red)(cancel(color(black)("L NaOH"))) × "0.500 mol NaOHl"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.0250 mol NaOH}$

The chemical equation is

$\text{HCl + NaOH → NaCl" + "H"_2"O}$

You have 0.0250 mol $\text{HCl}$ and 0.050 mol $\text{NaOH}$, so you have 0.0250 mol of reaction.

There are two heats involved in this process:

$\text{heat from reaction + heat to warm calorimeter = 0}$

$\textcolor{w h i t e}{m m m m m} {q}_{1} \textcolor{w h i t e}{m m m} + \textcolor{w h i t e}{m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m m m l} = 0$

color(white)(mmmm)nΔ_text(r)Hcolor(white)(mm) + color(white)(mmmll)C_text(tot)ΔT color(white)(mmmmm)= 0

In this problem,

$n \textcolor{w h i t e}{m l l} = \textcolor{w h i t e}{l} \text{0.0250 mol}$
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#

$\text{-1418 J"color(white)(l) + C_text(tot) × "3.2 °C} = 0$

${C}_{\textrm{\to t}} = \text{1418 J"/"3.2 °C" = "440 J/°C}$