Question

< Question 3 of 3 > Ammonia, NH, reacts with oxygen to form nitrogen gas and water. 4NH, (aq) + 30,(g) + 2N (8) + 6 H,0(1) If 2.45 g of NH, reacts with 3.68 g of O, and produces 0.850 L of N, at 295 K and 1.00 atm, which reactant is limiting? O NH3(aq) O 02(8) What is the percent yield of the reaction? percent yield: 55.7

Answer 1

4NH3 (aq) + 302 (9) -> 2N29) + 6H2O (1) moles of NH₃ = mass molar mass 2.45 = 17 0.1441 moles of NH₃ moles of Oz - o 3.68 = 0.115 males olo, moles of he needs motes A B 3 moles of O₂ requires 4 moles of NH3 1 mol of O₂ requires y moles of NH₃ 01115 0.115 mol of O2 requires = 4 x med moles of NH₃ 0.153 = cod moles of NH3 since requires moles (0.153) of NH3 is more than available (0.1441) moles of NH3 , so, NH₂ (aq) reactant is limiting. PV = nRT 1X0.850 = nx 0.0821 x 295 0.850 0.0821 X295 n = 0.0351 males of N2

PAGE 10 4 moles of NH3 ramive produce a moles of N2 I moles of NH3 produce ( 3 = 1) males of 0.1441 moles of NH3 produce = txo.1441 = 0.0720 moles of Na % yield x 100 yield actual yield theortical yield 0.0351 100 0.0720 48.75% Answer Uplo yield =