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 16.70 g of ethyl alcohol?

Answer 1

To calculate the enthalpy change for burning 16.70 g of ethyl alcohol, determine the number of moles based on its molar mass (46.07 g/mol) and then use that value to find the enthalpy change by multiplying with the given ΔH per mole (-1,367 kJ).

Step-by-step explanation:

The enthalpy change for the combustion of ethyl alcohol (also known as ethanol) can be calculated using stoichiometry based on the given thermochemical equation:

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

First, we need to determine the moles of ethyl alcohol in 16.70 g. The molar mass of C2H5OH is approximately 46.07 g/mol. Therefore, the number of moles is:

Number of moles = mass/molar mass = 16.70 g / 46.07 g/mol

Next, we can use this amount to find the enthalpy change by proportion:

ΔH (for 16.70 g) = -1,367 kJ * (number of moles of C2H5OH from above)

Finally, we calculate the enthalpy change for 16.70 g of ethyl alcohol by plugging in the number of moles into the equation.