Thermochemistry

 84) ΔH_c = -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% = m_ethanol/m_sol'n x 100%

       m_ethanol = 180 g x 0.106 = 19 g C₂H₅OH

       n_Cal = 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

       n_Cal = -1.3 x 10² Cal = 1.3 x 10² Cal