Answer:
Explanation:
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In this case, since the combustion analysis provides us the yielded mass of both water and carbon dioxide, as the sources of hydrogen and carbon in the hydrocarbon, we are able to compute the moles of each via the following mole-mass relationships:
Now, we divide those moles by the fewest ones (those of carbon) in order to compute their subscripts in the empirical formula:
However we need whole numbers, that is why we multiply each subscript by 4 to get 4 for carbon and 5.08 ≅ 5 for hydrogen:
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Answer:
Kc = 3.94
Explanation:
CH₃COOH(g) + C₂H₅OH(g) → CH₃COOC₂H₅(g) + H₂O(g)
Liquids aren't included in the equilibrium constant, but at 100°C, all four compounds are a gas. So the equilibrium equation is:
Kc = [CH₃COOC₂H₅] [H₂O] / ([CH₃COOH] [C₂H₅OH])
Set up an ICE table for each row and calculate the value of Kc. Then average the results. (See picture.)
Kc ≈ (3.919 + 4.012 + 3.902) / 3
Kc ≈ 3.94
