Mass of methane required to increase the temperature of the air in the room by 7.35 °C is 7.95 g
The volume of air in the room is first calculated:
Volume of air in the room = 4.33 m x 3.43 m x 3.03 = 45.00 m³
1 m³ = 1000 L
45.00 m³ = 45.00 m³ * 1000 L/m³
Volume of air in L = 45000 L
Number of moles of air in 45000 L of air is then determined:
1.00 moles of air occupies 22.4 L
number of moles of air in 45000 L = 45000 L * 1 mole / 22.4 L
number of moles of air = 2008.93 moles of air
Energy that is needed to heat the room by 7.35 °C is then calculated:
Quantity of energy needed = Specific heat capacity * number of moles * temperature increase
Specific heat capacity of air = 30.0 J/K/mole
Quantity of energy needed = 30.0 * 2008.93 * 7.35
Quantity of energy needed = 442969.065 J = 443.00 kJ
The amount of methane required to produce that amount of energy is then calculated:
Equation of combustion of methane : CH₄ + 2 O₂ ---> CO₂ + 2 H₂O
Enthalpy of combustion of methane = −890.3 kJ/mole
Number of moles of methane required = 443.00 kJ / 890.8 kJ/mole = 0.497 moles
Mass of 1 mole of methane = 16.0 g
mass of 0.497 moles of methane = 16.0 * 0.497 = 7.95 g
Therefore, mass of methane required to increase the temperature of the air in the room by 7.35 °C is 7.95 g
Learn more at: https: brainly.com/question/4213585
