Answer :
Answer:
0.92 kg
Explanation:
The volume occupied by the air is:
[tex]35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L[/tex]
The moles of air are:
[tex]3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol[/tex]
The heat required to heat the air by 10.0 °C (or 10.0 K) is:
[tex]1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J[/tex]
Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:
[tex]5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}[/tex]
Combustion is a type of chemical reaction in which a compound reacts in the presence of air to form the product. The mass of methane required in the reaction is 0.92 kg.
What is mass?
Mass is the amount of the weight or the matter that depicts the quantity in grams, kilograms etc.
The volume occupied by the air in the house is calculated as:
[tex]35 \times 35 \times 3.2 \times \dfrac{10^{3}\;\rm L}{1\;\rm m^{3}} = 3.9\times 10^{6}\;\rm L[/tex]
The number of moles of air is:
[tex]3.9\times 10^{6}\;\rm L\times \dfrac{1\;\rm mol}{22.4\;\rm L}= 1.7\times 10^{5}\;\rm mol[/tex]
The heat needed to heat the air of the room is:
[tex]1.7\times 10^{5}\;\rm mol \times \dfrac{30\;\rm J}{\rm K.mol}\times 10\;\rm K= 5.1\times 10^{7}\;\rm J[/tex]
The heat of combustion of methane is 55.5 MJ/kg so the mass of methane will be:
[tex]5.1\times 10^{7}\;\rm J \times \dfrac{1}{55.5\times 10^{6}}= 0.92\;\rm kg[/tex]
Therefore, 0.92 kg is the mass required of methane.
Learn more about combustion and mass here:
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