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Chapter 6: Problem 99

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? $\left(\Delta H_{\text { combustion }}^{\circ} \text { for } \mathrm{CH}_{4}=-891 \mathrm{kJ} / \mathrm{mol}\right)$

Short Answer

Expert verified
The volume of methane required to heat the home can be calculated using the given energy needed and the standard enthalpy change for the combustion of methane as follows: Volume of methane = \(\frac{4.19 \times 10^{6} kJ}{-891 kJ/mol}\) × \(22.4 L/mol\) By calculating the above expression, we find the required volume of methane that must be burned to provide the necessary energy to heat the home.

Step by step solution

01

Calculate the number of moles of methane needed

First, let's determine how many moles of methane are required to provide the given energy. We know that the energy needed is \(4.19 \times 10^{6} kJ\), and the standard enthalpy change for the combustion of methane is \(-891 kJ/mol\). To find the number of moles, we'll use the formula: Moles of methane = \(\frac{\text{Energy needed}}{\Delta H_{combustion}}\) Moles of methane = \(\frac{4.19 \times 10^{6} kJ}{-891 kJ/mol}\)
02

Calculate the volume of methane required at STP

Now, we need to convert the moles of methane into a volume, which we can do using the molar volume of a gas at STP. At STP (standard temperature and pressure), the molar volume of any gas is \(22.4 L/mol\). To find the volume of methane required, we'll use the formula: Volume of methane = Moles of methane × Molar volume at STP Volume of methane = \(\frac{4.19 \times 10^{6} kJ}{-891 kJ/mol}\) × \(22.4 L/mol\) Once you have calculated the volume of methane, you will have the required amount of methane that must be burned to heat the home with the given energy.

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Most popular questions from this chapter

Calculate \(\Delta H\) for the reaction: $$ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1010 . \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-317 \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-623 \mathrm{kJ}\) $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-286 \mathrm{kJ}\)

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