Question

The thermochemical equation for the burning of ethyl alcohol is

C₂H₅OH(l) + 3O₂(g) --> 2CO₂(g) + 3H₂O(l)    ΔH = -1,367 kJ

What is the enthalpy change (in kJ) for burning 10.01 g of ethyl alcohol?

Answer 1

We have given,

The thermochemical equation for the burning of ethyl alcohol is,

C₂H₅OH(l) + 3O₂(g) 2CO₂(g)+ 3H₂O(l)------- H = -1367 kJ

Thus, For 1 mole of C₂H₅OH = -1367 kJ enthalpy change

Calculating the moles for 10.01 g of ethyl alcohol,

= 10.01 g x ( 1 mole/46.07 g)

= 0.2173 moles of ethanol

Now,

= 0.2173 moles of ethanol x ( -1367 kJ / 1 mole)

= 297.0 kJ

Thus, 297.0 kJ of enthalpy change for burning 10.01 g of ethyl alcohol.