A) 382.70 g
B) 832.74 g
C) 463.9 g
D) 684.0 g
Correct Answer: D
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 \[=\frac{2700}{1352.9}=2\,mol\] \[=2\times 242=684g\]You need to login to perform this action.
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