Question

1.07 gg H2H2 is allowed to react with 9.59 gg N2N2, producing 1.75 gg NH3NH3.

A) What is the theoretical yield in grams for this reaction under the given conditions?

B) What is the percent yield for this reaction under the given conditions?

Answer 1

a for gm H2 t 9.59 gm N2 malar mass of H2 = 25 mol moles of Haz 1-07 = 0.535 moles molar mass of N2 = 28 shmal a mole of n2 = 9.59 = 0-343 moles 0-343 mox N2 + 3H2 - 2 NH3 be 0.535 mol 1 mol Na need 3 mol H₂ so a 0.343 mol N2 will need - 3x00343 = 1.03 mol H₂ but we only have 0-335 mol H₂ So Ha is limitting seagent - 3 mol H2 gives a mel NH3 So 0.533 mol H2 will give 7/8 x 0.535= 0.357 mol NH3. molar mass of NH3 =17 md. a theoritical sield of NH3 = 0.357x17 = 6.0 69m actual yield of NH3 = 1.75 gm op gield = 175 X 100 I (cs Scann le giedo = 28.9%) CamScanner