| N2(g) + 3H2 19 2NH2 (9) . This is an example of a Haber Process for the Synthesis Ammonia from Hydrogen and Nitrogen. 3 mol H2 Tomes based on the following calculation: 2 I mol H3 = 3 moles H2 1 2 moles NH3 Now for 834.8 mols of Nutty 2 moles NH3 = 58.2 Moles of H2 (c), 14P + 502-> 2205 0. Arronding back to balanced Reaction 5 moles a gives – 3 moles a gives 5 moles hos a moles pros 5 moles Oz 2 moles mol off 5 mol, mol

Ros forms moles at 1:36. For 5 moles or more, the number of moles required to form moles is 4 for I mole. Fox 30.1 moles are needed to make Qr i.l.u = p will. Moles of Allla Foormed = 2 for 1 Mole of Gillz = 2 Moles of Alles – for 6 moles Bulla = 2 xomolles of Alla = 2 moles Of Alleg – will Form

I mole of All too Number of moles Culla Regarred = 3-3 moles 1 Required for 4.5 moles in Alles – X9.5 = 6.75 Moles of Culla. – Reputable Maximum I have Miq Four questions can be answered. You can answer 5 questions for convenience. Thanks for asking question