Question

part a
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 10.0 °c. assume the following: house dimensions are 34.0 m×31.0 m×3.2 m ; molar heat capacity of air is 29 j k⁻¹ mol⁻¹ ; 1.00 mol of air occupies 22.8 l for all temperatures concerned. δch° for methane is - 802.5 kj mol⁻¹.
express your answer in grams to two significant figures.
m =
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Explanation:

Step1: Calculate the volume of the house

The volume $V$ of a rectangular - shaped house with dimensions $l\times w\times h$ is given by $V = l\times w\times h$. Here, $l = 34.0\ m$, $w = 31.0\ m$, and $h = 3.0\ m$. So, $V=34.0\times31.0\times3.0\ m^{3}=3198\ m^{3}$. Since $1\ m^{3}=1000\ L$, the volume of the house in liters is $V = 3198\times1000\ L=3.198\times 10^{6}\ L$.

Step2: Calculate the number of moles of air

We know that $1.00\ mol$ of air occupies $22.8\ L$. Let $n_{air}$ be the number of moles of air. Then $n_{air}=\frac{V}{22.8\ L/mol}=\frac{3.198\times 10^{6}\ L}{22.8\ L/mol}\approx1.4026\times 10^{5}\ mol$.

Step3: Calculate the heat required to heat the air

The heat capacity of air is $C = 29\ J\ K^{-1}\ mol^{-1}$, and the temperature change $\Delta T=10.0^{\circ}C = 10.0\ K$. The heat required to heat the air $q$ is given by the formula $q = n_{air}C\Delta T$. Substituting the values, we get $q=(1.4026\times 10^{5}\ mol)\times(29\ J\ K^{-1}\ mol^{-1})\times(10.0\ K)=4.06754\times 10^{7}\ J = 4.06754\times 10^{4}\ kJ$.

Step4: Calculate the number of moles of methane required

The enthalpy of combustion of methane $\Delta_{c}H^{\circ}=- 802.5\ kJ/mol$. Let $n_{CH_4}$ be the number of moles of methane. The heat released by the combustion of methane is equal to the heat required to heat the air. So, $|q| = n_{CH_4}|\Delta_{c}H^{\circ}|$. Then $n_{CH_4}=\frac{q}{|\Delta_{c}H^{\circ}|}=\frac{4.06754\times 10^{4}\ kJ}{802.5\ kJ/mol}\approx50.7\ mol$.

Step5: Calculate the 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 $m$ of methane is given by $m = n_{CH_4}M$. Substituting the values, we get $m=(50.7\ mol)\times(16.05\ g/mol)\approx814\ g\approx810\ g$ (to two significant figures).

Answer:

$810$