# Question #0e53e

##### 1 Answer

#### Answer:

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

#color(blue)(q = m * c * DeltaT)" "# , where

*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_"sol" = V_"HCl" + V_"NaOH"#

#V_"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. color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 11.2color(red)(cancel(color(black)(""^@"C")))#

#q = "1872.6 J"#

Now focus on the chemical equation for this neutralization reaction

#"HCl"_text((aq]) + "NaOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O"_text((l])#

Notice that you have a **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

#color(blue)(c = n/V implies n = c * V)#

#n_"HCl" = "3.0 M" * 20. * 10^(-3)"L" = "0.060 moles HCl"#

and

#n_"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

This means that the amount of heat absorbed by the water was released when

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

#color(blue)("% error" = (|"experimental value" - "theoretical value"|)/"theoretical value" xx 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.*