Chem 1320 1st Edition Lecture 11Outline of Last Lecture I. DilutionII. GasesIII. AssumptionsIV. Important VariablesOutline of Current Lecture I. EquationsII. ExamplesIII. Gas StoichiometryCurrent LectureI. EquationsFixn= n1=n2 = A1T1 = P1 = T1Fixv= v1=v1 A2T2 P2 T2n= PV/ RTP= nRT/ VPV= nRTII. Examples1. 16 grams of oxygen in a 5L vessel at 25 ˚C, what’s the pressure? V= 5L, T= 25 + 273.15= 298.15K, 16gO2= 0.5molO2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.P (5) = (0.5) (0.08206) (298.15)5 P= 2.45atm2. ˚What is the volume of 49.8g HCl at STP (standard temperature and pressure)? STP: T= 298.15K, P= 1atm n: 49.8g / 36.458= 1.366mol1(V) = (1.366mol) (0.08206) (298.15) = V= 33.42L3. What volume if CO2 at 37˚C and 1atm when 5.6g glucose is used? C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) T= 37˚C = 310.15K, P= 1atmn= 5.6g C6H12O6 1 mol C6H12O6 6CO2 180g 1 mol C6H12O6PV= nRT (l)(V) (0.187)(0.08206)(310.15)V= 4.76L4. A lightbulb has 1.2atm and 18˚C and is heated to 85˚C what is the final pressure? P= 1.2atm, T= 18˚C= 291.15K, P=?, T= 85˚C, C= 358.15K1.2 = 291.15X 358.15 P2= 1.48atmIII. Gas Stoichiometry1). Balanced chemical reaction- molar mass (solid)- molarity(liquid)- n= PV/RT (gas)2). Get known species into moles3). Relate unknown and known species with stoichiometry from a balanced equation4). Convert to what is
View Full Document
Unlocking...