We are asked to determine the mass of methane required to heat the air in a house by 15.0 ˚C.

Calculate mole air needed:

$\mathbf{mol}\mathbf{}\mathbf{air}\mathbf{}\mathbf{=}[\left(35.0m\times 35.0m\times 3.2m\right)\times {\left(\frac{1\mathrm{dm}}{{10}^{-1}m}\right)}^{3}\times \frac{1L}{1{\mathrm{dm}}^{3}}]\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}}{\mathbf{22}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{L}}\phantom{\rule{0ex}{0ex}}\mathbf{mol}\mathbf{}\mathbf{air}\mathbf{}\mathbf{=}[\left(3920\overline{){m}^{3}}\right)\times {\left(\frac{1\overline{)\mathrm{dm}}}{{10}^{-1}\overline{)m}}\right)}^{3}\times \frac{1\overline{)L}}{1\overline{){\mathrm{dm}}^{3}}}]\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}}{\mathbf{22}\mathbf{.}\mathbf{4}\mathbf{}\overline{)\mathbf{}\mathbf{L}}}$

mol air = 175,000 mol air

Calculate heat:

$\overline{){\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{nC}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{}\mathbf{=}(175,000\overline{)\mathrm{mol}})(30\frac{J}{\overline{)\u02daC}\u2022\overline{)\mathrm{mol}}})(15.0\overline{)\u02daC})$

**q= 7.875 x 10 ^{4 }kJ**

We'll use the thermochemical equation to calculate the mass of CH_{4}.

Molar mass CH_{4} = 16.05 g/mol

CH_{4}(g) + O_{2}(g) → CO_{2}(g) + H_{2}O(g); ΔH_{rxn}˚ = –802.3 kJ/mol

Combustion of natural gas (primarily methane) occurs in most household heaters. The heat given off in this reaction is used to raise the temperature of the air in the house. Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of methane required to heat the air in a house by 15.0 ˚C. Assume each of the following: house dimensions are 35.0m x 35.0m x 3.2 m; specific heat capacity of air is 30 J/K•mol; 1.00 mol of air occupies 22.4 L for all temperatures concerned.