# Chemistry - Limiting Reactants, Thermochemical Change

• Wolfowitz
In summary, the problem involves mixing a 14.1-mL sample of 0.996 M NaOH with 32.3 mL of 0.905 M HCl in a coffee-cup calorimeter and determining the final temperature of the reaction mixture. The enthalpy of the reaction is -55.8 kJ, and both solutions are at 21.6°C before mixing. The solutions manual gives a final temperature of 25.6°C, while the individual attempting to solve the problem gets 25.7°C. The difference in results may be due to the method used to determine the final temperature, with the solutions manual dividing the heat released by the specific heat capacity of the entire system, while the individual
Wolfowitz
The problem is that 'my' solution is at variance with the solution provided for the question in the solutions manual that accompanies the textbook I am using.

## Homework Statement

"A 14.1-mL sample of 0.996 M NaOH is mixed with
32.3 mL of 0.905 M HCl in a coffee-cup calorimeter (see Section
6.6 of your text for a description of a coffee-cup calorimeter).
The enthalpy of the reaction, written with the lowest
whole number coefficients, is 55.8 kJ. Both solutions are at
21.6°C prior to mixing and reacting. What is the final temperature
of the reaction mixture? When solving this problem, assume
that no heat is lost from the calorimeter to the surroundings,
the density of all solutions is 1.00 g/mL, the specific
heat of all solutions is the same as water, and volumes are

## Homework Equations

Besides the conversion factors,
q(absorbed) = -q(released)
q = specific heat capacity * mass * Δtemperature

## The Attempt at a Solution

The final temperature is the temperature recorded when thermochemical equilibrium is attained.

Given the 1.00 mL:1.00 g conversion, it means that the entire mass of the solutions is 46.4 g.

The thermochemical system is the NaCl and the H2O, which releases 55.8 kJ of heat when formed from 1 mol of NaOH and 1 mol of HCl.

The reaction is as follows: NaOH + HCl → NaCl + H2O: ΔH = -55.8 kJ.

The reaction is ushered in as the result of a mixing of 14.1mL of a 0.996M NaOH solution, or

14.1mL NaOH sol.$\frac{0.996 mols NaOH}{1000mL NaOH sol.}$ = 0.0140 mols NaOH,

and a 32.3mL HCl solution of 0.905M, or

32.3mL NaOH sol.$\frac{0.905 mols HCL}{1000mL HCl sol.}$ = 0.0292 mols HCl,

There is a limiting reactant here, and so given that 0.0140 mols NaOH reacts, the amount of heat released will be equivalent to (0.0140/1)(55.8 kJ) = 0.781 kJ or 781 J.

Easy stuff so far.

The formation of 0.0140 mols NaCl and 0.0140 mols H2O releases 781 J to the surroundings. By the equation listed above,

q(released)(-1) = q(absorbed)
781 J = (massunreacted solution)(4.18J/g*°C)(Tfinal - 21.6°C)
Tfinal = [781 J]/[(4.184J/g*°C)*massunreacted solution] + 21.6°C

If all the NaOH reacts with 0.0140 mols of HCl to form 0.0140 mol of NaCl and 0.0140 mol of H2O to form 1.08g of H2O and NaCl, which is part of the same thermochemical system, the only place the heat can go is to the unreacted solution of the HCl, this much of which remains:

Massunreacted solution = Mass of solutions before reaction - (Mass of H2O + Mass of NaCl) = (32.3 + 14.1)g - 1.08g = 45.3g.

Substituting, we get

Tfinal = $\frac{781 J}{45.3g*(4.18J/g*°C)}$ + 21.6°C
= 25.7°C

as the 'final temperature'.

The solutions manual, in deriving the answer, for Tfinal, divides 781 J by the specific heat capacity of the solution , the same as that of water (see the question), and the total mass of the solutions before reaction, 46.4 g. This doesn't seem correct to me. Accordingly, their solution is 25.6°C (not much of a difference). I'm mainly concerned about whether my method is correct, because to not discern between system and surroundings in such a context is to miss a big one.

So: Is my method correct?

Solutions manual is right. You can't heat up just part of the mixture, you heat up everything - so you heat up whole 46.4 g.

What is - in your version - final temperature of the 1.08 g of water produced?

## 1. What is a limiting reactant in chemistry?

In chemistry, a limiting reactant is a substance that is completely used up during a chemical reaction, limiting the amount of product that can be formed. It is the reactant that is present in the smallest quantity, and once it is used up, the reaction stops.

## 2. How do you determine the limiting reactant in a chemical reaction?

To determine the limiting reactant in a chemical reaction, you must first write out a balanced equation for the reaction. Then, you will need to calculate the amount of product that can be formed from each reactant. The reactant that produces the smallest amount of product is the limiting reactant.

## 3. What is the difference between an excess reactant and a limiting reactant?

An excess reactant is a reactant that is present in a larger quantity than necessary for the reaction to occur. It is not completely used up in the reaction. A limiting reactant, on the other hand, is completely used up and determines the amount of product that can be formed.

## 4. How does the concept of limiting reactants apply to real-world scenarios?

The concept of limiting reactants is important in many areas of chemistry, including industrial processes and food production. In these scenarios, it is crucial to know the limiting reactant in order to optimize the reaction and produce the desired amount of product efficiently.

## 5. What is thermochemical change in chemistry?

Thermochemical change is a process in which heat energy is either released or absorbed during a chemical reaction. This can result in a change in temperature, as well as the formation or breaking of chemical bonds. Thermochemical change is important in understanding the energy changes that occur during chemical reactions.

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