What is the energy given off with the following information? How do I figure this?

5.43 g of CH4 are burned in 7.23 g of O2. The % yield of the experiment was only 75.3%. How much energy was given off in this reaction?





1 CH4 + 2 O2鈫? CO2 + 2 H2O 鈭咹 = -890 kJ





What is the energy given off with the following information? How do I figure this?
The molar mass of CH4 is 16.0 g/mol; for O2 it is 32.0 g/mol.





5.43 g CH4 x (1 mol / 16.0 g) = 0.339 mol CH4


7.23 g O2 x (1 mol / 32.0 g) = 0.226 mol O2





Find the limiting reagent:


1 mol CH4 reacts with 2 mol O2


0.339 mol CH4 x (2 mol O2 / 1 mol CH4) = 0.678 mol O2 required to use up all the CH4. Since there is only 0.226 mol O2, it is the limiting reagent. Use O2 to determine the yield.





0.226 mol O2 x (2 mol H2O / 2 mol O2) = 0.226 mol H2O





% yield = (actual yield / theor. yield) x 100


75.3 = (actual yield / 0.226 mol) x 100


actual yield = 75.3 x 0.226 mol / 100 = 0.170 mol H2O





2 mol H2O formed releases 890 kJ.


0.170 mol H2O x (890 kJ / 2 mol) = 75.7 kJ of heat is given offWhat is the energy given off with the following information? How do I figure this?
5.43 g CH4 x (1 mol CH4 / 16 g) x (1 mol CO2 / 1 mol CH4) = 0.34 mol CO2





7.23 g O2 x (1 mol O2 / 32 g) x (1 mol CO2 / 2 mol O2) = 0.11 mol CO2





Since the Oxygen leads to a smaller amount of product, it is the oxygen that represents the limiting reactant.





7.23 g O2 x (1 mol O2 / 32 g) x (890 kJ / 2 mol O2) = 101 kJ of heat should be produced IF the reaction was running at 100% yield.





Since the yield was only 75.3 % the amount of heat given off should be





101 kJ x 0.753 = 76.1 kJ