The complete combustion of ethanol, C2H5OH (FW = 46.0 g/mol), proceeds as follows:

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = −555 kJ
What is the enthalpy change for combustion of 15.0 g of ethanol

-181.0

-181 kJ

-555 kJ x (15.0/46.0) = ?

181 Well, the enthalpy change for the combustion of 15.0 g of ethanol can be calculated using the equation:

ΔH = (n × ΔHf) / M

Where:
ΔH is the enthalpy change
n is the number of moles
ΔHf is the enthalpy of formation
M is the molar mass

First, let's find the number of moles of ethanol in 15.0 g:

n = m / M
n = 15.0 g / 46.0 g/mol
n ≈ 0.33 mol

Now, we can substitute the values into the equation:

ΔH = (0.33 mol × -555 kJ) / 1 mol
ΔH ≈ -183.15 kJ

So, the enthalpy change for the combustion of 15.0 g of ethanol is approximately -183.15 kJ. But hey, at least it's not on fire anymore!

To find the enthalpy change for the combustion of 15.0 g of ethanol, we can use the equation:

ΔH = (molar mass of ethanol) * (ΔH per mole of ethanol) * (moles of ethanol)

First, let's calculate the moles of ethanol using the given mass of 15.0 g and its molar mass.

Molar mass of ethanol (C2H5OH) = 46.0 g/mol

Moles of ethanol = (mass of ethanol) / (molar mass of ethanol)
= 15.0 g / 46.0 g/mol
≈ 0.3261 mol (rounded to four decimal places)

Now, we can calculate the enthalpy change using the given ΔH per mole of ethanol:

ΔH = (-555 kJ/mol) * (moles of ethanol)
= (-555 kJ/mol) * 0.3261 mol
≈ -180.8 kJ (rounded to one decimal place)

Therefore, the enthalpy change for the combustion of 15.0 g of ethanol is approximately -180.8 kJ.