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To solve the above question, we have to calculate the gram equivalent of both the solutions. We also have to know whether there is any limiting reagent or not?

A limiting reagent is a reactant in the solution that gets consumed in the chemical reaction and therefore limits how much product can form.

Reagent – 1

\(\begin{array}{l}{\rm{ No of equivalent = Volume of solution }} \times {\rm{ Molarity(M)}}\\{\rm{No of equivalent of }}100{\rm{ ml of 0}}{\rm{.250 M HCl solution = 100}} \times 0.250\\{\rm{ = 25}}\end{array}\)

Similarly for reagent – 2

\(\begin{array}{l}{\rm{No of equivalent = Volume of solution }} \times {\rm{ Molarity(M)}}\\{\rm{No of equivalent of 2}}00{\rm{ ml of 0}}{\rm{.150 M NaOH solution = 200}} \times 0.150\\{\rm{ = 30}}\end{array}\)

We can observe that no of equivalent is greater for NaOH, which means HCl will be the limiting reagent here.

From the above chemical reaction, we can say that on the complete reaction of 1 mol of HCl and 1 mol of NaOH, 58 kJ energy was released.

\(\begin{array}{l}{\rm{Here, limiting reagent is HCl}}\\{\rm{No}}{\rm{. of moles of HCl = Volume of HCl(in L) }} \times {\rm{ Molarity(M)}}\\{\rm{ }} = {\rm{ }}\frac{{100}}{{1000}} \times 0.250{\rm{ }}mol\\{\rm{ }} = {\rm{ }}0.0250{\rm{ }}mol\end{array}\)

To calculate the amount of energy released, we have to follow the given steps,

\(\begin{array}{l}{\rm{ }}1{\rm{ mol HCl releases 58 kJ of energy}}\\0.0250{\rm{ mol of HCl will release }}58 \times \frac{{0.0250}}{1}kJ{\rm{ energy}}\\{\rm{ = }}1.45{\rm{ k}}J{\rm{ energy}}\end{array}\)

Hence, the amount of energy released is 1.45 kJ.

To calculate the change in temperature of the solution, we have to assume that the density of the solution is \(1.00{\rm{ }}g/ml\)_.

First, we will calculate the mass of the solution from the given information about the volume of the solution.

\(\begin{array}{l}{\rm{Mass = Volume }} \times {\rm{ Density}}\\\therefore {\rm{ Mass of solution = Volume of solution (100 ml + }}{\rm{ 200 ml) }} \times {\rm{ Density of solution(1}}{\rm{.00 g/ml)}}\\{\rm{ = }}300{\rm{ }}ml{\rm{ }} \times {\rm{ }}1.00{\rm{ }}g/ml\\{\rm{ }} = {\rm{ }}300{\rm{ }}g\end{array}\)

Now, we will calculate the change in temperature using the given information about the heat capacity of the products and the previously calculated amount of heat released.

\(\begin{array}{l}{\rm{Heat, }}\Delta {\rm{q = mc}}\Delta {\rm{T}}\\{\rm{[Here, }}\Delta {\rm{q}},{\rm{ is change in energy, }}\\{\rm{ m is mass of the solution}}\\{\rm{ c is the heat capacity of the solution}}\\{\rm{ }}\Delta {\rm{T is the change in temperature}}\end{array}\)

\(\begin{array}{l}{\rm{Here, }}\\\Delta {\rm{q }} = - 1.45{\rm{ }}KJ{\rm{ }} = - 1450{\rm{ }}J\\{\rm{and m = 300 g}}\\{\rm{and c = }}4.19{\rm{ }}J/g{\rm{ }}^\circ C\\ \Rightarrow \Delta {\rm{q}} = {\rm{mc}}\Delta {\rm{T}}\\ \Rightarrow - 1450J = (300g) \times (4.19J/g{\rm{ }}^\circ C) \times \Delta {\rm{T}}\\ \Rightarrow \Delta {\rm{T}} = - 1.15^\circ C\end{array}\)

Hence, the temperature change will be \(1.15^\circ C\)in the solution. We have to assume the density of the solution, 1.00 g/ml throughout the solution.