You are planning a lecture demonstration to illustrate an endothermic process. You want to lower the temperature of \(1400 \mathrm{mL}\) water in an insulated container from 25 to \(10^{\circ} \mathrm{C} .\) Approximately what mass of \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) should you dissolve in the water to achieve this result? The heat of solution of \(\mathrm{NH}_{4} \mathrm{Cl}\) is \(+14.7 \mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{Cl}\).

Short Answer

Expert verified
Approximately 316g of ammonium chloride need to be dissolved in the water to achieve the desired result.

Step by step solution

01

Understand the given data and requirements

The given data include the initial volume of water (\(1400 \mathrm{mL}\) or \(1.4 \mathrm{L}\)), the initial temperature (25°C), the final temperature (10°C), and the heat of solution for \(\mathrm{NH}_{4} \mathrm{Cl}\) (+14.7 kJ/mol). The task is to find the mass of the \(\mathrm{NH}_{4} \mathrm{Cl}\) needed to reach the final temperature.
02

Calculate the required heat to cool the water

Calculate the heat (q) required to cool the water using the formula q = m*c*ΔT, where m is the mass of the water (in kg), c is the specific heat capacity of water (4.18 kJ/kg°C), and ΔT is the change in temperature (in °C). The mass of the water can be calculated knowing that 1L of water is approximately equivalent to 1 kg, thus making the mass 1.4 kg. Therefore q = 1.4 kg * 4.18 kJ/kg°C * (25 - 10) °C = 87.36 kJ.
03

Calculate the moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) needed

The moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) required can be calculated using the formula q = n*ΔH, where n is the number of moles and ΔH is the heat of solution. Therefore n = q / ΔH = 87.36 kJ / 14.8 kJ/mol = 5.90 mol.
04

Find the mass of \(\mathrm{NH}_{4} \mathrm{Cl}\)

The molar mass of \(\mathrm{NH}_{4} \mathrm{Cl}\) is approximately 53.49 g/mol. Therefore, the mass can be found by multiplying the number of moles by the molar mass: Mass = n * molar mass = 5.90 mol * 53.49 g/mol = 315.59 g.

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Most popular questions from this chapter

Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction.Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction. (a) \(0.584 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) yields \(29.4 \mathrm{kJ}\) (b) \(0.136 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(\mathrm{s}),\) yields \(5.27 \mathrm{kJ}\) (c) \(2.35 \mathrm{mL}\) of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(\mathrm{l})(d=0.791\) \(\mathrm{g} / \mathrm{mL}),\) yields \(58.3 \mathrm{kJ}\)

For the reaction \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)\) determine \(\Delta H^{\circ},\) given that $$\begin{array}{r} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-202.4 \mathrm{kJ} \end{array}$$ $$\begin{aligned} 2 \mathrm{HCl}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \\ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)+\mathrm{H}_{2} \mathrm{O}(1) & \Delta H^{\circ}=-318.7 \mathrm{kJ} \end{aligned}$$

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

A 0.75 g sample of \(\mathrm{KCl}\) is added to \(35.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) in a Styrofoam cup and stirred until it dissolves. The temperature of the solution drops from 24.8 to \(23.6^{\circ} \mathrm{C}\) (a) Is the process endothermic or exothermic? (b) What is the heat of solution of KCl expressed in kilojoules per mole of KCl?

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).

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