# Question 0e53e

Nov 12, 2015

Here's what I got.

#### Explanation:

The idea here is that the heat released by the neutralization reaction will be absorbed by the solution, hence the recorded increase in temperature.

You can assume that the solution behaves similar to pure water, since the values given for specific heat and density are consistent with those used for water.

So, the equation that establishes a relationship between heat absorbed and change in temperature looks like this

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

Now, the volume of the solution will be equal to the sum of the volumes of the hydrochloric acid and sodium hydroxide solutions

${V}_{\text{sol" = V_"HCl" + V_"NaOH}}$

${V}_{\text{sol" = "20. mL" + "20. mL" = "40. mL}}$

Use the density of the solution to find its mass

40. color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "40. g"

The heat absorbed by the water will thus be

$q = 40. \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 11.2color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{1872.6 J}$

Now focus on the chemical equation for this neutralization reaction

${\text{HCl"_text((aq]) + "NaOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have a $1 : 1$ mole ratio between hydrochloric acid and sodium hydroxide. This means that the reaction will consume equal numbers of moles of each reactant.

Use the molarities and volumes of the two solutions to determine how many moles of each you have

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{\text{HCl" = "3.0 M" * 20. * 10^(-3)"L" = "0.060 moles HCl}}$

and

${n}_{\text{NaOH" = "2.0 M" * 20. * 10^(-3)"L" = "0.040 moles NaOH}}$

Since you have fewer moles of sodium hydroxide, the base will act as a limiting reagent, i.e. it will determine how many moles of hydrochloric acid actually take part in the reaction.

More specifically, the base will be completely consumed by the reaction, and only $0.040$ moles of hydrochloric acid will react.

This means that the amount of heat absorbed by the water was released when $0.040$ moles of each reactant took part in the reaction.

The heat given off by this reaction per mole will thus be

1color(red)(cancel(color(black)("mole"))) * "1872.6 kJ"/(0.040color(red)(cancel(color(black)("moles")))) = "46815 J"

The enthalpy change of neutralization will be

DeltaH_"neut" = color(green)(-"47 kJ/mol") -> rounded to two sig figs

The negative sign symbolizes heat given off.

Percent error is defined as

$\textcolor{b l u e}{\text{% error" = (|"experimental value" - "theoretical value"|)/"theoretical value} \times 100}$

You can discard the minus signs for this calculation

"% error" = ( |47 - 55.9|color(red)(cancel(color(black)("kJ/mol"))))/(55.9color(red)(cancel(color(black)("kJ/mol")))) xx 100 = color(green)(16%)#

SIDE NOTE The experimental value you got for this neutralization reaction is actually characteristic of a weak acid reacting with a strong base, so I would definitely say that significant errors were introduced in the experiment.