Question

For the following reaction, 6.25 grams of nitrogen gas are mixed with excess hydrogen gas . Assume that the percent yield of ammonia is 89.9 %.

nitrogen(g) + hydrogen(g) -> ammonia(g)

What is the ideal yield of ammonia ?	grams
What is the actual yield of ammonia ?	grams

Answer 1

mais of Nq = 6.25 g no. of male of N2 = 6.25g 28glmal = 0.2232 mol. Neeg, + 3H2(g) → 2NH3 (9) I male Na produce a mad NH₃, then 0.2232 maal Ne produe 2x0.2232 = 0.4464 nad NH₂ Selee_0 Чrecel of = очч64 mоѕ0 х 17, = 7.599 Hence, 7.59 g es ideal yield of ammonia. % yield = 89.9% = 89.9 Actual yield xwo Ideal yield 89.9 x 7.59 Actual yield - 100 = 6.829 Hence, 6.82g es actual yield of NH₃.