A bomb calorimeter is used to determine the amount of energy liberated when a sa

Question

A bomb calorimeter is used to determine the amount of energy liberated when a sample of propane is combusted. The bomb calorimeter is calibrated by combusting 7.77 g of methane (CH4, MW=16.0 g/mol, Delta Ecomb =-902k3/mol) which leads to a temperature change of 40.8 degreeC. After the calibration of the bomb you load 3.25 g of propane into the same calibrated calorimeter and perform your combustion experiment. During this experiment, there is a recorded increase of temperature of 15.2 degree C. What is the deltaE of combustion for propane (MW = 44.1 g/mol). Please report your answer to the correct number of significant figures. kJ/mol

Explanation / Answer

given

7.77 g of methane

we know that

moles = mass / molar mass

so

moles of methane = 7.77 / 16

so

moles of methane = 0.485625

now

heat of combustion of methane = 902 kJ/mol

now

heat evolved = moles of methane x heat of combustion of methane

so

heat evolved = 0.485625 x 902

heat evolved = 438.03375

now

for calorimeter

heat evolved = heat capacity x temp change

so

438.03375 = heat capacity x 40.8

heat capacity = 10.73612132 kJ/ C

now

for propane

heat evolved = heat capacity x temp change

heat evolved= 10.73612132 x 15.2

heat evolved = 163.1890441 kJ

given

3.25 g of propane

so

moles of propane = 3.25 / 44.1

moles of propane =

so

heat of combustion of propane = heat evolved / moles of propane

so

heat of combustion of propane = 163.1890441 / ( 3.25 / 44.1)

heat of combustion of propane = 2214.35 kJ/mol

so

the heat of combustion of propane is -2214.35 kJ/mol

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