1. The problem statement, all variables and given/known data A power plant burns oil that is 4% ash and 3% Sulfur. At 50% excess air, what particulate (mg/m^3) emissions would you expect? 2. Relevant equations PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil Start with 1000 g of oil oil = 4% ash, 3%sulfur, and 93% carbon oxygen is 20% of air PM will be found in units of mg/m^3 PV = nrt where p = 1 atm, r= .08206 l*atm/mole*k, t= 298K 3. The attempt at a solution I started with the simple combustion equation provided by the professor: C + O2 ---> CO2 The mole ratio of each part is 1. I then found the amount in grams or carbon. Since we started with 1000g of oil, we would have 930g of Carbon. 930g C x (1 mole carbon/12g of carbon)x(1 moleO2/1mole C) = 77.5 moles of carbon. Since the mole ratio was one to one. Then there would be 77.5 moles of oxygen as well. 77.5 moles of carbon x (1 mole of O2/ 1 mole of C) = 77.5 moles of O2. Then found the amount of O2 burned intially. 77.5 moles of O2 X (32g of O2/1 mole of O2) = 2480 g of O2. Now to find the moles of air used. 77.5 moles of O2 = .20 x (moles of air) x = 387.5 moles of air Now that the moles have been found, you can find the volume of air using Pv=nrt v=nrt/p v= ((387.5 moles of air)*(.08206 L*atm/mol*K)*(273K))/ 1atm v= 8680.92 L air was in 50% excess in the reaction so there needs to be a conversion. 8680.92 L x ( 1.5)= 13021.4 L of air volume needs to be in m^3 13021.4 L x (1 m^3/1000 L) = 13.021 m^3 The amount of PM is equal to: 2.8 x .04 = .112 grams of PM This needs to be in mg .112 grams of PM x (1000 mg/ 1 gram) = 112 grams of PM Final solution of PM: 112 grams of PM / 13.021 m^3 = 8.60 mg/m^3 of PM ***** I am not sure if i correctly interpreted the excess air part of the formula. Is my method correct for finding moles of O2 to then find moles of air and using the ideal gas law? Thanks.