Many portable gas heaters and grills use propane, c3h8.
using enthalpies of formation, calculate the quantity of heat produced when 10.0 g of propane is completely combusted in air under standard conditions. assume that liquid water is forming. express your answer using four significant figures.

503.5 kJ

Explanation:

The combustion of reaction of propane, C3H8, is:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)

The enthalpy of the reaction can be calculated by the enthalpy of formation of the substances, which is the enthalpy of the reaction that produces the substances by only its constituents. The values can be found at a thermodynamic table. The standard condition is 25°C and 1 atm, so:

H°f, C3H8(g) = -103.85 kJ/mol

H°f, O2(g) = 0

H°f, CO2(g) = -393.51 kJ/mol

H°f, H2O(l) = -285.83 kJ/mol

So, the enthalpy of the reaction is:

ΔH°rxn = ∑n*H°f products - ∑n*H°f reactants, where n is the coefficient of the substance:

ΔH°rxn = [3*(-393.51) + 4*(-285.83)] - (-103.85)

ΔH°rxn = -2220 kJ/ mol of C3H8

The heat produced is the value of the enthalpy multiplied by the number of moles of the fuel. The molar mass of C3H8 is 44.10 g/mol, so, in 10.0 g:

n = 10.0/44.10

n = 0.2268 mol

So, the heat is:

Q = -2220 * 0.2268

Q = -503.5 kJ

The minus signal indicates that the heat is being lost, so 503.5 kJ of heat is produced.

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