Question

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 the following: house dimensions are 31.0 m × 34.0 m × 2.8 m ; molar heat capacity of air is 31 j k⁻¹ mol⁻¹ ; 1.00 mol of air occupies 22.2 l for all temperatures concerned. δch° for methane is - 802.5 kj mol⁻¹. express your answer in grams to two significant figures.

Explanation:

Step1: Calculate volume of the house

$V = 31.0\ m\times34.0\ m\times2.8\ m=2951.2\ m^{3}$
Since $1\ m^{3}=1000\ L$, then $V = 2951200\ L$.

Step2: Determine moles of air

Number of moles of air, $n_{air}=\frac{2951200\ L}{22.2\ L/mol}=132936.94\ mol$

Step3: Calculate heat required to heat the air

The heat - capacity formula is $q = nC\Delta T$. Here, $C = 31\ J\ K^{-1}\ mol^{-1}$, $\Delta T=15.0\ ^{\circ}C = 15.0\ K$ (since a change in Celsius is equal to a change in Kelvin).
$q=n_{air}C\Delta T=132936.94\ mol\times31\ J\ K^{-1}\ mol^{-1}\times15.0\ K = 6.188\times10^{7}\ J=61880\ kJ$

Step4: Calculate moles of methane

The heat of combustion of methane is $\Delta_{c}H^{\circ}=- 802.5\ kJ/mol$. Let the number of moles of methane be $n_{CH_{4}}$.
We know that the heat released by methane is equal to the heat required to heat the air. So, $n_{CH_{4}}=\frac{61880\ kJ}{802.5\ kJ/mol}=77.11\ mol$

Step5: Calculate mass of methane

The molar mass of methane ($CH_{4}$) is $M=(12.01 + 4\times1.01)\ g/mol = 16.05\ g/mol$
The mass of methane, $m = n_{CH_{4}}\times M=77.11\ mol\times16.05\ g/mol = 1237.62\ g\approx1200\ g$ (to two significant figures)

Answer:

$1200\ g$