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Chapter 6: Problem 70

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.

Short Answer

Expert verified
The final temperature of the mixture when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.100 M Ba(OH)₂ is approximately 27.12°C.

Step by step solution

01

Calculate the limiting reagent

Determine the limiting reagent by calculating the moles of HCl and Ba(OH)₂ in the given volumes and concentrations of each solution. Moles of HCl: moles = Molarity × Volume moles = 0.500 M × 0.100 L = 0.050 moles Moles of Ba(OH)₂: moles = Molarity × Volume moles = 0.100 M × 0.300 L = 0.030 moles Based on the stoichiometry of the reaction (2 moles of HCl react with 1 mole of Ba(OH)₂), twice the moles of HCl are needed in comparison to the moles of Ba(OH)₂. But in this case, there are more moles of HCl present than required, making Ba(OH)₂ the limiting reagent.
02

Calculate the heat evolved in the reaction

Using the moles of the limiting reactant (Ba(OH)₂), the enthalpy of the reaction (ΔH), and the stoichiometry of the reaction, calculate the heat evolved (q) during the reaction. q (reaction) = Moles of limiting reactant × ΔH q (reaction) = 0.030 moles Ba(OH)₂ × (-118000 J/mol) q (reaction) = -3540 J Notice that the result is negative, which indicates that the system has released heat (exothermic reaction).
03

Calculate the heat absorbed by the solution

Now, determine the heat absorbed by the solution (q(solution)), which is equal in magnitude but opposite in sign to the heat evolved in the reaction (q(reaction)). Calculate the temperature change in the solution using this heat absorbed and the specific heat capacity provided. q (solution) = -q (reaction) q (solution) = 3540 J ΔT (temperature change) = q(solution) / (mass × specific heat capacity) ΔT = 3540 J / (400.0 g × 4.18 J/g·°C) ΔT = 2.12°C
04

Calculate the final temperature of the mixture

Using the initial temperature (25.0°C) and the temperature change (ΔT) calculated in the previous step, determine the final temperature of the mixture. Final temperature = Initial temperature + ΔT Final temperature = 25.0°C + 2.12°C Final temperature = 27.12°C Now we have determined that when mixing 100 mL of 0.500 M HCl with 300 mL of 0.100 M Ba(OH)₂, the final temperature of the resulting solution is approximately 27.12°C.

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Most popular questions from this chapter

A 5.00 -g sample of aluminum pellets (specific heat capacity \(=\) 0.89 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity $=0.45 \mathrm{J} / \mathrm{C} \cdot \mathrm{g}\( are heated to \)100.0^{\circ} \mathrm{C}$ . The mixture of hot iron and aluminum is then dropped into 97.3 \(\mathrm{g}\) water at $22.0^{\circ} \mathrm{C}$ . Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

A biology experiment requires the preparation of a water bath at $37.0^{\circ} \mathrm{C}$ (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C},\) and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C} .\) If a student starts with 90.0 \(\mathrm{g}\) cold water, what mass of hot water must be added to reach $37.0^{\circ} \mathrm{C} ?$

In a coffee-cup calorimeter, 50.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{AgNO}_{3}\) and 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\) , and the final temperature is \(23.40^{\circ} \mathrm{C}\) Calculate the heat that accompanies this reacture in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \rho \mathrm{C} \cdot \mathrm{g} .\)

Consider the reaction $$ 2 \mathrm{ClF}_{3}(g)+2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+6 \mathrm{HF}(g)+\mathrm{Cl}_{2}(g)\quad\Delta H^{\circ}=-1196 \mathrm{kJ} $$ Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{ClF}_{3}(g)\)

A piston performs work of \(210 . \mathrm{L} \cdot\) atm on the surroundings, while the cylinder in which it is placed expands from \(10 . \mathrm{L}\) to 25 \(\mathrm{L}\) . At the same time, 45 \(\mathrm{J}\) of heat is transferred from the surroundings to the system. Against what pressure was the piston working?
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