$$ \ce{C6H5OH(s) + 7O2(g) -> 6CO2 + 3H2O(l)} $$

When a \(2.000\)-gram sample of pure phenol, \(\ce{C6H5OH(s)}\), is completely burned according to the equation above, \(64.98\) kilojoules of heat is released. Calculuate the molar heat of combustion of phenol in kilojoules per mole.

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We need to find the kilojoules of heat released per mole of phenol:

$$ \frac{\pu{64.98 kJ}}{\pu{2.000 g }\ce{C6H5OH(s)}} \cdot \frac{\pu{94 g }\ce{C6H5OH(s)}}{\pu{1 mol }\ce{C6H5OH(s)}}$$ $$ \Delta H_{\text{comb}}^{\circ} = \pu{3,054 kJ/mol} $$

Because this heat is being released, the sign will be negative.