# I need some thermodynamics help.What mass of propane (C3H8 (g)) must be burned to supply 2775 kJ of heat? The standard enthalpy of combustion of propane at 298 K is -2220 kJ*mol-1.

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In order to produce the number of mass of propane needed we have to follow the steps. First divide the heat needed by the enthalpy of combustion of propane which is -2220 kJ/mol. This time disregard the negative sign first.

2775 kJ /2220 kJ/mol = 1.25 moles propane

Next, obtain the mass of propane by multiplying the moles produced with the molar mass of propane which is 44.1 g/mol

1.25 mol x 44.1 g/mol = 55.125 grams of propane

**55.13 grams** of propane is needed in order to produce 2775 kJ of heat.