Thermochemistry

 

 84) ΔHc = -1367 kJ/mol            D = 1.0 g/mL          1 cal = 4.18 J

       m%ethanol = 10.6%          V = 177 mL

 

       C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)     ΔH = -1367 kJ

 

       D = M/V

       M = D x V = 1.0 g/mL x 177 mL = 180 g sol'n

 

       m% = methanol/msol'n x 100%

       methanol = 180 g x 0.106 = 19 g C2H5OH

 

       nCal = 19 g C2H5OH x 1 mol C2H5OH/46.08 g C2H5OH x -1367 kJ/1 mol C2H5OH x 1 cal/4.18 J x 103 J/1 kJ x 1 Cal/103 cal

       nCal = -1.3 x 102 Cal = 1.3 x 102 Cal