Question
Asked Nov 3, 2019

If all of the energy from burning 294.0 g of propane (ΔΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of water can be heated from 20.0°C to 100.0°C? (Assume that the density of water is 1.00 g/mL)

 

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Step 1

Given information:

Mass of propane = 294.0 g

Initial temperature = 20.0 ˚C

Final temperature = 100.0 ˚C

Density of water = 1.00 g/mL

Enthalpy of combustion (ΔcombH) = -2220 kJ/mol

Step 2

First, number of moles of propane are calculated as follows:

294.0 g
44.0956 g/mol
6.667 mol
= 6.67 mol
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294.0 g 44.0956 g/mol 6.667 mol = 6.67 mol

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Step 3

The amount of energy produced by 6.67 moles ...

Amount of energy = 6.67 mol 2220 kJ/mol
14807.4 kJ
14807400 J
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Amount of energy = 6.67 mol 2220 kJ/mol 14807.4 kJ 14807400 J

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