Question

17. A 3.00 mole sample of gas contains oxygen, nitrogen, and chlorine. If the partial pressures of the three gases are in a 2:3:7 ratio respectively, how many grams of nitrogen are present in the sample? 1) 14 grams 2) 21 grams 3) 28 grams 4) 49 grams

Answer 1

Solutions we know that From ideal gas law [PEORT] Fъя ous" ач (о») where Por is partial pressure of 02. 102 is mole of O2 (RT) is constant. Same as For nitrogen gas (2) Porz = har (RI) Para = nora (27) For chiorine gas (ce) [See = "ce (2) En Pee = nce We also know that - fatal = fatal (PT) Where Pratal is pressure of all gases e tatal is tatal male of gases. (RT) is constant

From 0 0.0 and Porxh taral Po₂ = 102 x Pratal Protal =- tatal no 2 PN2 = 2 x Ptatal Protal = Prax tatal • DN2 total Pe= re & Protal » Pratal = Peexntatal tatal he So, . Portatal no2 Para tatal: o2 Pe xn total = x nee Where k= constant given (partial pressure of e ga Pos: Pora: Pce 2:3: 7 total = tatal male = 3 mol So De pozPure Pace = constante 2 = 3 - 7 = constant no2 ON2 non Na ne Ho2 = nye hice = content mole of chlorine of 2 - - mole of O2 mole e of o2 = 2 - 3

mole Xtoutal male 12 male. so the male ratio of Oz ös: ce is (2:3:7 male of N2 = 3 x total male = 3x3 = 3 male I male of 2 = 20g so 3 male of N2= 2 x 20g = 219 ale Anso sol Nitrogen present in sample = 21 grams