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?.

Answers

Answer 1

The enthalpy change for combustion of 15.0 g of ethanol in the chemical reaction is -180.93 kJ.

The enthalpy change for combustion of 15.0 g of ethanol can be calculated using the equation:

ΔH = (ΔH°/mol) × (mol/mass) × mass

Where ΔH° is the standard enthalpy change for the reaction, mol is the number of moles of ethanol, and mass is the mass of ethanol in grams.

Given that the standard enthalpy change for the combustion of ethanol is -555 kJ, the molecular weight of ethanol is 46.0 g/mol, and the mass of ethanol is 15.0 g, we can plug these values into the equation:

ΔH = (-555 kJ/mol) × (1 mol/46.0 g) × (15.0 g)

Simplifying the equation gives:

ΔH = (-555 kJ/mol) × (0.326 mol)

ΔH = -180.93 kJ

Therefore, the enthalpy change for combustion of 15.0 g of ethanol is -180.93 kJ.

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