Question

Solid ammonium chloride, NH, CI, is formed by the reaction of gaseous ammonia, NH, and hydrogen chloride, HCI. NH,(6) + HCl(g) NH,Cl(s) A 3.45 g sample of NH, gas and a 3.45 g sample of HCl gas are mixed in a 0.50 L flask at 25°C. Identify the limiting reagent OHCI ONH, O NH, C1 How many grams of NH, CI will be formed by this reaction? mas What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH CI produced by the reaction

Answer 1

Therefore from 3.45g of HU 335 x 3.45 - sosg of Nhy I will form Remaining NH3 - (5.05-1.6079 - 3.459 For NH3, Moles = = Given Mass Molar Mass 3.45 17 = 6.202 nol only gas remaining in the NH₃ is the flask, Therefore the pressure of the Gas is, PV = nRT P= nRT Pe ? n=0.202 mol R=0.082 Latmmol/k T = 25° c = 2988 V=0.50L P = 0.202 x 0. 082x 298 0.50 Pa 9.87 atru

Actual Molas matio of NHS: HUL = 0.202 : 0.094 2.14 : 1 → NH Therefore, NH₃ is in excess amount and Hill is the limiting agent. So, Hll is the limiting agent & NH₃ + HU NHU U a mole of Hll reacts with Imole of NH₃ 36.5g of He react with ing of NH3 3.45g of Hell react with 17 X 3.45 - 3605 -1.60g of NH3 from 36.5g of Hel 53.5g of NHy ll.

I Given, Mass of NH3 gas = 3.45g Mass of Hill gas = 3.45g Moles of NH₃ and Hell can be calculated as, Molas mass of NH2 = 14 + 3x1 = 17g | neol Molar Mass of Hel= Molar mass of H = 1+ 35.5= 36.5 g 1 mol Moles of NH = Given May = 3.45g = 0.202 g mol Molar Mass 10 Moles of H = Given Mass Molar Mass - 3.45a 36.50 = 0.094 mol Inol The balanced reaction is given as: NH₃(g) + Hulg) Nhy Cl (9) Stoichiometric Molar ratio of NH : HCl =1:1