# 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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### 1 Answer

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.