What is the enthalpy change when 4.608 g of ethanol, C2H5OH(ℓ), undergoes complete combustion?

Question

C2H5OH(ℓ) + 3O2(g) --> 2CO2(g) + 3H2O(ℓ) + 13 668 kJ

my answer:
N (C2H5OH) = m/M
n = 4.608g / 46.08 g /mol
n= 0.1000 mol

the question said to use this formula: ▲H = n▲H°
but can someone show me how to use this formula

Answer 1

I can't show you but I can tell you since you don't have the dHo values shown. Look up in tables (those in your book or use Google). You want to find the dHformation values for each of the products and each of the reactants.Then
dHrxn = (n*dH of products) - (n*dH of reactants)
n is the coefficients in the reaction. It's 1 for example for ethanol and 3 for O2.
It's really pretty simple once you obtain the dH values from a table.
When you find dHrxn that will be for the combustion of 1 mol ethanol. You have a mass ethanol given and it isn't 1 mole. Calculate how many moles that is then convert from dH rxn for 1 mol to dHrxn for the moles in the problem. Good luck. Show your work if you get stuck.

Answer 2

delta H=mols x delta H (comb)
delta H=0.1 x 13668