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- Frage: (thermo-102)
Beim Umsatz von 60 Gramm Traubenzucker werden x kJ frei. Berechne daraus die ΔHof.
Textantwort: M(Traubenzucker) = M(C6H12O6) = 180 Gramm
Mit 60 g C6H12O6 x kJ, bei einem Mol (=180 g) somit 3·x.
ΔHof(C6H12O6) = 3·x kJ/mol.
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