In this example problem, we'll calculate the mass of the reactant needed based on the percent efficiency of a combustion reaction and the actual amount of heat that needs to be absorbed. Our problem reads: propane gas, which is C3H8, can be used as a fuel in a camping stove. What mass of propane is needed to cook a dinner if the dinner must absorb 990 kilojoules of heat and the camping stove has an efficiency of twelve percent? Our reaction for the combustion of propane gas is: C3H8 + 502 ⟶ 3 CO2 + 4 H2O. The change in enthalpy for the reaction is -2200 kilojoules. Our first step will be to calculate the theoretical amount of heat that we need. We can use our percent efficiency equation, where percent efficiency equals the actual amount of heat that we need divided by the theoretical amount of heat we need, times one hundred percent. We will rearrange this equation for the theoretical amount of heat, which is equal to the actual amount of heat divided by the percent efficiency, times one hundred percent. Our dinner must actually absorb 990 kilojoules of heat to cook, and our percent efficiency is twelve percent. Plugging these numbers into our equation, we should get 8.3 × 10^3 kilojoules of heat. This is the amount that the dinner must theoretically absorb at twelve percent efficiency. Next, we'll calculate the mass of the reactants we need to be able to produce this theoretical amount of heat. The heat released by the reaction is equal to the negative of the heat absorbed by the dinner. So, we'll start our dimensional analysis to calculate the heat released by the reaction as -8.3 × 10^3 kilojoules. We will convert this amount of heat that must be released into a mass of propane gas using dimensional analysis. We can convert from kilojoules into moles of propane using the change in enthalpy given in the equation and our coefficient for propane in the balanced chemical equation. Then, we can convert from moles of propane into grams of propane using its molar mass. Completing this calculation, we will get 170 grams of propane. So, by burning 170 grams of propane, our dinner should be able to absorb 990 kilojoules of heat, taking into account an efficiency of only twelve percent.