Question

Need help with part B

One possible use for the cooking fat left over after making french fries is to burn it as fuel. Formula = C_51 H_88 O_6 Density = 0.94g/mL Delta H^o_f = - 1310kJ/mol Write a balanced equation of the combustion of cooking fat. C_51 H_88 O_6(l) + 70 O_2 (g) rightarrow 51 CO_2(g) + 44 H_2 O(l) C_51 H_88 O_6(l) + 70 O_2 (g) rightarrow 7 CO_2(g) + 44 H_2 CO_3(l) C_51 H_88 O_6(l) + 48 O_2 (g) rightarrow 51 CO_2(g) + 44 H_2(l) C_51 H_88 O_6(l) + 70 O_2 (g) rightarrow 51 HCOOH(g) + H_2 O(l) Part B Use the data above to calculate the amount of energy released (in kilojoules per milliliter) from the combustion of cooking fat: Express your answer using two significant figures. q = ________ kJ/mL

Answer 1

A) Balnced equation for combustion of cooking fat is

C51H88O6 (l) + 70O2 (g)----------> 51CO2 (g) + 44H2O (l)

B) C51H88O6 (l) + 70O2 (g)----------> 51CO2 (g) + 44H2O (l)

    Given that H^of [C51H88O6(l)] = -1310 kJ/mol

         We know that    H^of [O2 (g)] = 0 kJ/mol

                                H^of [CO2 (g)] = -393.5 kJ/mol

                               H^of [H2O (l)] = -285.8 kJ/mol

ΔH^orxn = ΔH_f^o(products) - ΔH_f^o(reactants)

           = 51 x -393.5 kJ/mol + [ 44 x -285.8 kJ/mol] - [-1310 kJ/mol + 70 x 0 ]

          = -31333.7 kJ/mol

ΔH^orxn = -31333.7 kJ/mol

Negative sign indicates , heat energy is released.

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Molar mass of C51H88O6 = 796 g

   density = 0.94 g/mL

   Then,

   volume = mass/ density = 796 g / 0.94 g/mL = 846.8 mL

Heat energy released per mL

     = 31333.7 kJ / 846.8 mL

    = 37.0 kJ/mL

Therefore,

Heat energy released per mL = 37.0 kJ/mL