Question

Q3. (20 pts) Ammonium nitrite NH.NO2, decomposes according to the following chemical equation. NH.NO, (s) → N, (g) + 2H,0 (g) What is the total volume of products obtained when 128 g NH.NO, decomposes at 819 °C and 2.00 atm ? (Atomic weight of N=14.00 g/mol, H=1.01g/mol, O= 16.00 g/mol)

Answer 1

Total volume of products obtained = 268.72 L

Complete solution is shown below.

Given reaction NH4NO2(8) N₂ (g) + 2H₂O(g) According to the above reaction, 1 mol N2 1 mol NH4NO2 forms 2 mol H₂O 3 128g 1 mol NH4NO2 gives Total moles (Product) 3 moles of product Given, Amount of NHA NO2 molar mass of NHANO2 =Massof N+Axm. mass of Ht M. Massogn + 2x M. masiofo 14.00gmol +4x4.019 Imol + 14.00g/mal + 2x 16.00 gimol 64.04 g/mol :. moles of NH4NO2 Amount ling) 64.04 g/mol 1.998 mol 1288 Molar mass = So, total moes of products = 3x 1.998 mal = 5.994 moll Griven; Temperature = 819°C - 819 + 273.15 K 1092.15 K Pressure ;P 2.00 atm Gas constant; R=0.0821 L-atm/k-mol Number of moles; n = 5.994 mal

Using, PV = nRT 2.00 atm XV = 5.994 mol X 0.0821 l-am K-mol X1092.15 2V = 537.45 L V = 537.45L 2 V = 268.721