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The combustion reaction occurred in a closed room containing 5.56×104g of air originally at 21.7°C . Assume that all of the heat produced by the reaction was absorbed by the air (specific heat = 1.005 J/(g°C)) in the room.

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Answer:

The description for the given question is described in the explanation section below.

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

[tex]=0.383 \ mol\times -1270 \ kJ/mol[/tex]

[tex]=-486 \ kJ[/tex]

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