Given that #DeltaH_"combustion"^@# for #"propane"# is #-2043*kJ*mol^-1#, what mass of propane is required to give #425*kJ#?

Question

Wyatt Atkins · Accepted Answer

You have the equation; treat the energy as a product!

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g) + 2043# #kJ# Normally, we would write the amount of energy given as a negative number, but in this case, it is written as a reaction product, which it is. One mole of propane burns to produce 2043 kJ of energy, or 2043 kJ of energy evolved per mole of reaction as written! So if 425 kJ of energy are evolved, then #(425*kJ)/(2043*kJ)# #~= 1/5#. Thus, approximately, 1/5 of a mole of propane must be combusted with stoichiometric oxygen to evolve this amount of energy. You can calculate the molar quantity of propane in grams, and you need approx. 1/5 of this.

Mateo Aguilar · Answer

undefined To find the mass of propane required to give 425 kJ of energy, we can use the equation:

mass = energy / ΔH_combustion

mass = 425 kJ / (-2043 kJ/mol)

mass ≈ -0.208 mol

Since mass cannot be negative, this indicates an error in the calculation. The mass of propane required to give 425 kJ of energy cannot be determined with the given information because the result is negative, indicating an error in the calculation.