According to the equation below, what is the enthalpy change when 400.0 g of propane is burned in excess oxygen? c3h8 (g) + 5o2 (g) ) ↔ 3co2 (g) + 4h2o (l) δh = -2221kj

The enthalpy of reaction is already given to be -2221 kJ per mole of propane reacted. So, first let's determine the number of moles of propane. The molar mass of propane is 44 g/mol.

400 g C₃H₈ (1 mol/44g) = 9.091 moles

Thus, the enthalpy change is

ΔH = -2,221 kJ/mol * 9.091 mol = -20,191.11 kJ