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9.19 - opposed to how it works on paper In this case the...

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Key: 19. What is the percent yield of the combustion of 0.60 g of propanol (C 3 H 7 OH) to form carbon dioxide if the heat from the reaction is just enough to warm up 1 L of water by 2 ˚C ? Δ H (delta H) = -2000 kJ/ mol C 3 H 7 OH a. < 1 % b. 4.0 % c. 22 % d. 40 % e. 92 % f. There is not enough heat given off g. none of these Let’s think about what we need to really determine in this problem. We are told we need to determine the percent yield, which can be determined like this: Percent yield = Actual yield/ theoretical yield X 100%. We really need to deduce the actual yield and the theoretical yield. But the yield of what? Well, it doesn’t really matter as long as it is the actual and theoretical yield are talking about the same product. We could choose any of the products of the combustion reaction, including energy produced. Since using energy will be the simplest, lets do that. The actual yield of something is how much is actually produced during an experiment as
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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. Let’s 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 │2˚C = 8000 J │ 1kJ = 8kJ released = -8kJ 1g˚C │ │ 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 Δ 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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