Question

Ammonia, NH, can react with oxygen to form nitrogen gas and water 4 NH, (aq) +30,(g) -2N,(g) + 6 H,0(1) If 3.85 g NH3 reacts with 5.78 g O2 and produces 0.550 L N,, at 295 K and 1.01 bar, which reactant is limiting? NH, (aq) (3)o O Calculate the percent yield of the reaction. percent yield:

Answer 1

AZ - 4NH3 (22) + 302 (y) - 2N, (4) + (4011) gen! Ny-3.359 02 – 5-789 weight N .STOL 1-29562 p rolbar mole - weight holu najs weight of Nl3 = 3.85, nolar mass of NH3 = 17 g/mole Mole of WW3 = 3.25 = 0.22 mot I weight of o, 5.788 nolar mass of 9 = 32 Il mot mole of or = 5.78 = 0.18 mole genehod from geaction staictionetry the role of NH3 required 3 role of or for copliting I nole of Nit reque'and 3 role of oz. ten, for 0.22 role of Nng gequired= 32 80.22 role of =10.165]02 but mole of 0, (0.18) is excess, why is limiting reactant

Product calculation is done wring limiting recetaret 4 nole of Nhz - 2 nole of N2 I role of why & role of N2 then, for onde mole of Nithy 1 + xo.880) role of Nq - boll mole of Na

from ideal gas equation 2 PV = nRT. 8-314x10 1.0/x 0.550-nl n=0.023] Actal yield theoretical yield X Percent yield & ( /W to 0.023x10 I Percent yidel- 20.80% ]