Calculate ΔH when 2 moles of ethanol (C2H5OH) reacts with excess oxygen according to the following thermochemical equation?C2H5OH + 3O2 → 2CO2 + 3H2O ΔH = −1366.7 kJ

–2733.4 KJ

Explanation:

The balanced equation for the reaction is given below:

C₂H₅OH + 3O₂ —> 2CO₂ + 3H₂O

ΔH = −1366.7 kJ

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Finally, we shall determine the enthalpy change (ΔH) produced by the reaction of 2 moles of C₂H₅OH. This can be obtained as follow:

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Therefore, 2 moles of C₂H₅OH will react to produce enthalpy change (ΔH) of = 2 × −1366.7 = –2733.4 KJ.

Thus, enthalpy change (ΔH) obtained is –2733.4 KJ

answer; /// i do believe that the correct answer is (a) boiling

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.