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

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

delta H=mols x delta H (comb)

delta H=0.1 x 13668

Well, if you forgot how to use that formula, don't worry! I'm here to clown around and help you out.

To use the formula ▲H = n▲H°, you just need to plug in the values correctly.

In this case:
- n represents the number of moles of ethanol. You already calculated it to be 0.1000 mol.
- ▲H° represents the standard enthalpy change, which is given in the balanced chemical equation as 13,668 kJ.

So, just multiply the number of moles of ethanol by the standard enthalpy change:
▲H = (0.1000 mol)(13,668 kJ)

Now, multiply those two numbers together, and you'll get your answer!

Sure! To use the formula ▲H = n▲H°, you need to know the number of moles (n) and the standard enthalpy change (▲H°) for the reaction.

In this case, you have already calculated the number of moles of ethanol (C2H5OH). It is 0.1000 mol.

The standard enthalpy change (▲H°) for the combustion reaction of ethanol is given as 13,668 kJ.

To find the enthalpy change for the combustion of 4.608 g of ethanol, you can substitute the values into the formula:

▲H = n▲H°
▲H = (0.1000 mol)(13,668 kJ)

Now, you just need to multiply the number of moles by the standard enthalpy change:

▲H = 1.3668 kJ

Therefore, the enthalpy change for the complete combustion of 4.608 g of ethanol is 1.3668 kJ.

Sure! In the given equation, the stoichiometric coefficient of ethanol (C2H5OH) is 1. This means that one mole of ethanol will undergo combustion and produce a change in enthalpy of ΔH° = -13,668 kJ.

To calculate the enthalpy change (∆H) for 4.608 g of ethanol, you first need to determine the number of moles of ethanol (n). To do this, use the formula:

n = mass/molar mass

The molar mass of ethanol (C2H5OH) is calculated as follows:

Molar mass (C2H5OH) = 2 * Atomic mass (C) + 6 * Atomic mass (H) + 1 * Atomic mass (O)
= 2 * 12.01 g/mol + 6 * 1.01 g/mol + 1 * 16.00 g/mol
= 46.08 g/mol

Now, substitute the known values into the equation to calculate the number of moles of ethanol:

n = 4.608 g / 46.08 g/mol
n = 0.1 mol

Finally, use the formula:

ΔH = n * ΔH°

Substitute the values:

ΔH = 0.1 mol * (-13,668 kJ/mol)
ΔH = -1366.8 kJ

Therefore, the enthalpy change when 4.608 g of ethanol undergoes complete combustion is approximately -1366.8 kJ.

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.