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**Hint:**We can calculate the amount of heat evolved during a combustion process by calculating the enthalpy change of formation for every molecule involved in the chemical reaction for 1 mole using the formulas of density and moles.

**Formula used:**\[\Delta H_{rxn}^o = [6\Delta H(C{O_2}) + 3\Delta H({H_2}O)] - [\Delta H({C_6}{H_6}) + \dfrac{{15}}{2}\Delta H({O_2})]\] , \[density = \dfrac{{mass}}{{volume}}\] and \[moles = \dfrac{{given\,\,mass}}{{molar\,\,mass}}\]

**Complete step-by-step answer:**

Combustion of benzene results in evolution of heat and the following reaction occurs during the process.

\[{C_6}{H_6} + \dfrac{{15}}{2}{O_2} \to 3{H_2}O + 6C{O_2}\]

Let us now calculate the amount of benzene from the density provided in the question i.e. from the formula of \[density = \dfrac{{mass}}{{volume}}\] , we can calculate mass by the product of density and volume. Therefore, mass of benzene will be 0.87gm/ml \[ \times 100ml = 87gm\] and we know the molar mass of benzene is 78 gm. Now using these values, we can calculate the moles of benzene present in 100ml.

\[moles = \dfrac{{given\,\,mass}}{{molar\,\,mass}}\]

\[moles = \dfrac{{87}}{{78}} = 1.115moles\]

We will now calculate \[\Delta H\] of formation for each compound present. And as we know that enthalpy of formation is zero for molecules present in standard form, so the enthalpy of formation of oxygen will be zero.

We are given that for 18 grams of carbon or graphite, the energy evolved is 590KJ. Bit we have carbon of molar mass 12 gms, so for 12 grams of carbon, the energy evolved will be \[ - 590 \times \dfrac{{12}}{{18}} = - 393.33KJ/mol\].

Next, we are provided that 15889 KJ heat is required to dissociate all the molecules of 1 litre or 1000ml water. We know the density of water is 1gm/ml so using the formula, \[density = \dfrac{{mass}}{{volume}}\] we calculate the mass of water in 1000ml as density multiplied by volume i.e. 1000gms.

We can further determine the number of moles as the given mass is 1000gms and molar mass of water is 18gms. So, putting these values in \[moles = \dfrac{{given\,\,mass}}{{molar\,\,mass}}\] , we get

Moles = \[\dfrac{{1000}}{{18}}\] so, for these many moles of water, we have 15889KJ of energy. Then for 1 mole of hydrogen and oxygen will produce 1 mole of water and for that, we will have \[15889 \times \dfrac{{18}}{{1000}} = - 286KJ/mol\] energy released during its formation.

We are also provided the heat of formation of liquid benzene is 50 KJ/mol. Now let write the reactions for the calculated enthalpy of formation.

\[

C(s) + {O_2}(g) \to C{O_2}(g),\Delta H = - 393.33KJ/mol \\

2{H_2}O(l) \to 2{H_2}(g) + {O_2}(g),\Delta H = - 286KJ/mol \\

6C(s) + 3{H_2}(g) \to {C_6}{H_6}(l),\Delta H = 50KJ/mol \\

\]

Heat of reaction is the subtraction from sum of heat of formation of products to sum of heat of formation of reactants. Therefore, for this reaction, we can write the heat of reaction as:

\[

\Delta H_{rxn}^o = [6\Delta H(C{O_2}) + 3\Delta H({H_2}O)] - [\Delta H({C_6}{H_6}) + \dfrac{{15}}{2}\Delta H({O_2})] \\

= [6( - 393.3KJ) + 3( - 286KJ)] - [50KJ + \dfrac{{15}}{2}(0KJ)] \\

= - 3267KJ/mol \\

\]

**Therefore, we got the information that 78gm of benzene evolves 3267.8KJ amount of heat. So, 87gm of benzene will evolve \[ - \dfrac{{3267KJ}}{{78}} \times 87 = - 3642.8KJ\] .**

**Note:**The standard conditions of 298K temperature and 1 bar pressure are applicable for any enthalpy change such that it will be zero at STP. Even for standard enthalpy of combustion as it is not the combustion taking place at 298K because when a substance burns, its temperature is higher than 298K but it is the heat change that we look for.

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