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Chapter 7: Problem 28

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) gas \(\left(d=1.83 \mathrm{kg} / \mathrm{m}^{3}\right)\) is used in most gas grills. What volume (in liters) of propane is needed to generate \(273.8 \mathrm{kJ}\) of heat? $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.9 \mathrm{kJ} \end{array}$$

Short Answer

Expert verified
The volume of propane needed to generate 273.8 kJ of heat is \[\frac{\left(\frac{273.8}{2219.9} \times 44.09\right)}{1.83} \, \mathrm{L}\]

Step by step solution

01

Calculate the Moles of Propane Needed

First, identify the amount of heat needed, which is given as \(273.8 \, \mathrm{kJ}\). We also have the heat of combustion of propane \(\left(\Delta H^{\circ}\right) = -2219.9 \, \mathrm{kJ}\). The number of moles of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) required can be calculated using the equation: \[\mathrm{n} = \frac{\mathrm{Heat} \, \mathrm{needed}}{\mathrm{Heat} \, \mathrm{generated} \, \mathrm{by} \, \mathrm{one} \, \mathrm{mole} \, \mathrm{of} \, \mathrm{Propane}}\] Substituting the given values gives: \[\mathrm{n} = \frac{273.8}{2219.9} \, \mathrm{mol}\]
02

Calculate the Mass of Propane Needed

The mass of propane needed (\(\mathrm{m}\)) can be calculated from the number of moles and the molecular weight of propane. The molecular weight of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) is \(3(12.01) + 8(1.01) = 44.09 \, \mathrm{g/mol}\). Hence the mass needed can be calculated using the equation: \[\mathrm{m} = \mathrm{n} \times \mathrm{molar} \, \mathrm{mass}\] Substituting the values gives: \[\mathrm{m} = \frac{273.8}{2219.9} \times 44.09\]
03

Calculate the Volume of Propane Needed

The volume of propane needed (\(\mathrm{V}\)) can be calculated from the mass and the given density using the equation: \[\mathrm{V} = \frac{\mathrm{m}}{\mathrm{Density}}\] Substituting the values gives: \[\mathrm{V} = \frac{\left(\frac{273.8}{2219.9} \times 44.09\right)}{1.83} \, \mathrm{L}\]

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Most popular questions from this chapter

A 0.205 g pellet of potassium hydroxide, \(\mathrm{KOH}\), is added to \(55.9 \mathrm{g}\) water in a Styrofoam coffee cup. The water temperature rises from 23.5 to \(24.4^{\circ} \mathrm{C}\). [Assume that the specific heat of dilute \(\mathrm{KOH}(aq)\) is the same as that of water.] (a) What is the approximate heat of solution of \(\mathrm{KOH}\) expressed as kilojoules per mole of \(\mathrm{KOH}?\) (b) How could the precision of this measurement be improved without modifying the apparatus?

Can a chemical compound have a standard enthalpy of formation of zero? If so, how likely is this to occur? Explain.

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

In each of the following processes, is any work done when the reaction is carried out at constant pressure in a vessel open to the atmosphere? If so, is work done by the reacting system or on it? (a) Neutralization of \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) by \(\mathrm{HCl}(\mathrm{aq}) ;\) (b) conversion of gaseous nitrogen dioxide to gaseous dinitrogen tetroxide; (c) decomposition of calcium carbonate to calcium oxide and carbon dioxide gas.

In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.
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