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Chapter 7: Problem 64

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 \(0.500 \mathrm{M}\) 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 \(4.18 \mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g},\) calculate the final temperature of the mixture.

Short Answer

Expert verified
The final temperature of the mixture is approximately \(22.88\,^{\circ}\mathrm{C}\).

Step by step solution

01

Determine the limiting reactant

First, we need to determine the limiting reactant in this chemical reaction. Moles of reaction = moles (Volume × Molarity) HCl: $$100.0\,\mathrm{mL} * 0.5\,\mathrm{M} = 50.0\,\mathrm{mmol}$$ Ba(OH)2: $$300.0\,\mathrm{mL} * 0.1\,\mathrm{M} = 30.0\,\mathrm{mmol}$$ Calculate the required ratio of reactants to complete the reaction: $$2\,\mathrm{HCl} \longleftrightarrow 1\,\mathrm{Ba(OH)_2}$$ Required ratio: $$50\,\mathrm{mmol HCl} / 30\,\mathrm{mmol Ba(OH)_2} = 1.67$$ Since the required ratio (1.67) is greater than the actual ratio (2), \(\mathrm{Ba(OH)_2}\) is the limiting reactant.
02

Calculate the heat produced

Now, we'll calculate the heat produced by the reaction: Given: $$\Delta H = -118\,\mathrm{kJ/mol}$$ Calculate the heat produced by the reaction: $$q = moles \times \Delta H$$ $$q = 30\,\mathrm{mmol} \times -118\,\mathrm{kJ/mol} = -3.54 \times 10^3 \,\mathrm{J}$$
03

Calculate the temperature change

Next, we have to calculate the temperature change using the heat and specific heat capacity (Cp). Given: - Mass (m) of the mixture: \(400.0\,\mathrm{g}\) - Specific heat capacity (Cp): \(4.18\,\mathrm{J/g°C}\) Calculate the temperature change: $$q = m \times Cp \times \Delta T$$ $$\Delta T = \dfrac{q}{m \times Cp}$$ $$\Delta T = \dfrac{-3.54 \times 10^3 \,\mathrm{J}}{400.0\,\mathrm{g}\times 4.18\,\mathrm{J/g°C}} = -2.12\,^{\circ}\mathrm{C}$$
04

Calculate the final temperature

Finally, let's calculate the final temperature of the mixture. Given the initial temperature: $$T_{initial} = 25.0\,^{\circ}\mathrm{C}$$ Calculate the final temperature: $$T_{final} = T_{initial} + \Delta T $$ $$T_{final} = 25.0\,^{\circ}\mathrm{C} - 2.12\,^{\circ}\mathrm{C} = 22.88\,^{\circ}\mathrm{C}$$ The final temperature of the mixture is approximately \(22.88\,^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

On Easter Sunday, April \(3,1983,\) nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: $$2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$$ a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \%\) HNO \(_{3}\) by mass with a density of \(1.42 \mathrm{g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? \((\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{kJ} / \mathrm{mol})\) b. According to The Denver Post for April \(4,1983,\) authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\) what was their major concern?

Objects placed together eventually reach the same temperature. When you go into a room and touch a piece of metal in that room, it feels colder than a piece of plastic. Explain.

Given the following data $$\begin{array}{cl}\mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(g) & \Delta H=-1225.6 \mathrm{kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2967.3 \mathrm{kJ} \\\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & \Delta H=-84.2 \mathrm{kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & \Delta H=-285.7 \mathrm{kJ} \end{array}$$ calculate \(\Delta H\) for the reaction $$\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g)$$

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) \(30.0 \mathrm{L})\) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V:\) Calculate the work (in units of J) associated with the two pathways. Is work a state function? Explain.

A piston performs work of \(210 . \mathrm{L}\). \(\mathrm{atm}\) 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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