13. You want to heat the air in your house with natural gas (CH4). Assume that your house has 275 mJof floor area and that the ceilings are(2.50 m from the floors. The air in the house has a molar heat capacity of 29.1 J/(mole °C). The number of moles of air in the house can be found by assuming that the average molar mass of air is 28.9 g/mole and that the density of air at these temperatures is 1.22 g/L. What mass of methane do you have to burn to heat the air from 15.0 °C to 22.0 °C? Hint: first write the complete balanced equation for the combustion of methane (CH4), the reaction used to heat your house, and use the following information to obtain AHrxn for this process. Compound AH (kJ/mole) 1.229 -74.87 -393.509 CHA(g) Density air = (3)0) H2O(g) -241.83 CH4(g) + 2 O2(g) - CO2(g) + 2 H20(g) Using the above heats of formation, AHxn =-802.30 kJ/mole rxn Volume of room = 687.5 m 215 x 2.5 = 687,5m %3D 687.5 m 100° cm 1 mL Volume conversion = = 6.875x10 Liters 1 m 1cm* 1000 mL 1. Using volume, density, and molar mass, we obtain = 29022.5 moles of air The total heat required to warm the room from 15.0 °C to 22.0 °C is: (q = 5.912x10 kJ Using the quantity of heat needed to warm the house (5.912x10 kJ), along with the heat released from the combustion of methane (-802.3 kJ/mole), we obtain: (118 grams of CH4)required for combustion a9,022.5 molee ain a8.9g
13. You want to heat the air in your house with natural gas (CH4). Assume that your house has 275 mJof floor area and that the ceilings are(2.50 m from the floors. The air in the house has a molar heat capacity of 29.1 J/(mole °C). The number of moles of air in the house can be found by assuming that the average molar mass of air is 28.9 g/mole and that the density of air at these temperatures is 1.22 g/L. What mass of methane do you have to burn to heat the air from 15.0 °C to 22.0 °C? Hint: first write the complete balanced equation for the combustion of methane (CH4), the reaction used to heat your house, and use the following information to obtain AHrxn for this process. Compound AH (kJ/mole) 1.229 -74.87 -393.509 CHA(g) Density air = (3)0) H2O(g) -241.83 CH4(g) + 2 O2(g) - CO2(g) + 2 H20(g) Using the above heats of formation, AHxn =-802.30 kJ/mole rxn Volume of room = 687.5 m 215 x 2.5 = 687,5m %3D 687.5 m 100° cm 1 mL Volume conversion = = 6.875x10 Liters 1 m 1cm* 1000 mL 1. Using volume, density, and molar mass, we obtain = 29022.5 moles of air The total heat required to warm the room from 15.0 °C to 22.0 °C is: (q = 5.912x10 kJ Using the quantity of heat needed to warm the house (5.912x10 kJ), along with the heat released from the combustion of methane (-802.3 kJ/mole), we obtain: (118 grams of CH4)required for combustion a9,022.5 molee ain a8.9g
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter4: Energy And Chemical Reactions
Section: Chapter Questions
Problem 87QRT
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How do we get the numbers circled in yellow. (total heat required to heat the room from 15 to 22, and amount of grams required for combustion.)
Transcribed Image Text:13. You want to heat the air in your house with natural gas (CH4). Assume that your house has
275 mJof floor area and that the ceilings are(2.50 m from the floors. The air in the house
has a molar heat capacity of 29.1 J/(mole °C). The number of moles of air in the house can
be found by assuming that the average molar mass of air is 28.9 g/mole and that the density
of air at these temperatures is 1.22 g/L. What mass of methane do you have to burn to heat
the air from 15.0 °C to 22.0 °C? Hint: first write the complete balanced equation for the
combustion of methane (CH4), the reaction used to heat your house, and use the following
information to obtain AHrxn for this process.
Compound
AH: (kJ/mole)
(8)"H
(8)0)
(8)0°H
-74.87
-393.509
Density air = 1-229
-241.83
CH4(g) + 202(g) → CO2(g) + 2 H2O(g)
→ CO2(g) + 2 H20(g)
->
+ (3)*H
Using the above heats of formation, AHpxn =-802.30 kJ/mole rxn
UXI
Volume of room = 687.5 m 215x 2,5 = 687.5m²
%3D
687.5 m 100° cm
3
1 mL
Volume conversion =
=%3D6.875x10° Liters
1m3
1 cm3
1000 mL
1.
Using volume, density, and molar mass, we obtain = 29022.5 moles of air
The total heat required to warm the room from 15.0 °C to 22.0 °C is: (q = 5.912x10 kJ
%3D
Using the quantity of heat needed to warm the house (5.912x10 kJ), along with the heat released
from the combustion of methane (-802.3 kJ/mole), we obtain:
118 grams of CH4) required for combustion
a8.9g
99,022.5 moleo ain
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