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Unformatted text preview: opposed to how it works on paper. In this case, the actual yield of energy will be based on the experimental data provided. Lets do that first. The problem says that when 0.60 g propanol actually combusts, it heats up 1L of water by 2 C. We can determine the heat energy put into the water using the specific heat of water like so: 4J 1000g water 2C = 8000 J 1kJ = 8kJ released = -8kJ 1gC 1000J The theoretical yield in this case is how much heat should be given off during the combustion reaction if everything were to react perfectly. We can use the H value provided and do a simple dimensional analysis step to determine the ideal yield of energy:-2000 kJ 1mol C 3 H 7 OH 0.6g C 3 H 7 OH = -20kJ 1 mol C 3 H 7 OH 60g C 3 H 7 OH Now we just need to plug in our values to the percent yield equation to get:-8kJ/-20kJ X100% = 40%...
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This document was uploaded on 11/04/2011 for the course CHEM 106 at BYU.
- Fall '08