1. Tardigrade
  2. Question
  3. Chemistry
  4. The heat of formation of C12H22O11(s),CO2(g) and H2O(l) are -530,-94.3 and -68.3 kcal mol-1 respectively. The amount of C12H22O11 to supply 2700 kcal of energy is

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Q. The heat of formation of $ {{C}_{12}}{{H}_{22}}{{O}_{11}}(s),C{{O}_{2}}(g) $ and $ {{H}_{2}}O(l) $ are $ -530,-94.3 $ and $ -68.3 $ kcal $ mo{{l}^{-1}} $ respectively. The amount of $ {{C}_{12}}{{H}_{22}}{{O}_{11}} $ to supply 2700 kcal of energy is

MGIMS WardhaMGIMS Wardha 2013

Solution:

$ {{C}_{12}}{{H}_{22}}{{O}_{11}}(s)+12{{O}_{2}}(g)\xrightarrow[{}]{{}}12C{{O}_{2}}(g)+11{{H}_{2}}O(l) $ $ \Delta H_{comb}^{o}=[12\Delta H_{f}^{o}(C{{O}_{2}})]+11\Delta H_{f}^{o}({{H}_{2}}O] $ $ -\Delta H_{f}^{o}({{C}_{12}}{{H}_{22}}{{O}_{11}})] $ $ =-1352.9\text{ }kcal $ Thus, number of moles of $ {{C}_{12}}{{H}_{22}}{{O}_{11}} $ for getting 2700 kcal of heat $ =\frac{2700}{1352.9}=2\,mol $ $ =2\times 242=684g $