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Chapter 6: Problem 91

Assume that \(4.19 \times 10^{6} \mathrm{~kJ}\) of energy is needed to heat a home. If this energy is derived from the combustion of methane \(\left(\mathrm{CH}_{4}\right)\), what volume of methane, measured at STP, must be burned? \(\left(\Delta H_{\text {combustion }}^{\circ}\right.\) for \(\mathrm{CH}_{4}=-891 \mathrm{~kJ} / \mathrm{mol}\) )

Short Answer

Expert verified
The volume of methane required to provide the given amount of energy to heat the home is approximately 105,484.56 liters at STP.

Step by step solution

01

Calculate moles of methane needed to produce the given energy

We are given the energy needed to heat the home as 4.19 x 10^6 kJ and the enthalpy of combustion of methane as -891 kJ/mol. We will use these values to find the moles of methane needed to produce the given energy. Energy_needed = Enthalpy_of_combustion x Moles_of_methane Moles_of_methane = Energy_needed / Enthalpy_of_combustion = \( \frac{4.19 \times 10^6 \,\text{kJ}}{-891 \,\text{kJ/mol}} \)
02

Solve for moles of methane

Now we divide the energy needed by the enthalpy of combustion to find the moles of methane needed: Moles_of_methane = \( \frac{4.19 \times 10^6}{-891} \) Moles_of_methane ≈ 4702.47 mol
03

Use the ideal gas law to find the volume of methane

Now, we have the moles of methane needed. We will use the ideal gas law, PV=nRT, to find the volume of methane at STP (Standard Temperature and Pressure). At STP, T = 273.15 K and P = 1 atm (1 atm = 101325 Pa). R (universal gas constant) = 8.314 J/(mol*K) Since pressure is given in atm, we will use R in L*atm/mol*K which is equal to 0.0821 L*atm/mol*K. Now we can solve for the volume of methane using the ideal gas law, PV=nRT: V = \( \frac{nRT}{P} \) V = \( \frac{4702.47 \, \text{mol} \times 0.0821\, \text{L*atm/mol*K} \times 273.15\, \text{K}}{1 \, \text{atm}} \)
04

Solve for the volume of methane

Now we can calculate the volume of methane required: V ≈ \( \frac{4702.47 \times 0.0821 \times 273.15}{1} \) V ≈ 105484.56 L The volume of methane required to provide the given amount of energy to heat the home is approximately 105,484.56 liters at STP.

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Most popular questions from this chapter

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

What is incomplete combustion of fossil fuels? Why can this be a problem?

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Given the following data $$ \begin{aligned} \mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \Delta H &=-393.5 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

The heat capacity of a bomb calorimeter was determined by burning \(6.79 \mathrm{~g}\) methane (energy of combustion \(=-802 \mathrm{~kJ} / \mathrm{mol} \mathrm{CH}_{4}\) ) in the bomb. The temperature changed by \(10.8^{\circ} \mathrm{C}\). a. What is the heat capacity of the bomb? b. A \(12.6-\mathrm{g}\) sample of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), produced a temperature increase of \(16.9^{\circ} \mathrm{C}\) in the same calorimeter. What is the energy of combustion of acetylene (in \(\mathrm{kJ} / \mathrm{mol}\) )?
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