The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g)3H2(g)+N2(g)→2N H3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation.1.29g H2 is allowed to react with 9.55g N2, producing 1.49g NH3.
What is the theoretical yield for this reaction under the given conditions?
What is the percent yield for this reaction under the given conditions?

Challenge question: this question is worth 6 points. as you saw in problem 9 we can have species bound to a central metal ion. these species are called ligands. in the past we have assumed all the d orbitals in some species are degenerate; however, they often are not. sometimes the ligands bound to a central metal cation can split the d orbitals. that is, some of the d orbitals will be at a lower energy state than others. ligands that have the ability to cause this splitting are called strong field ligands, cnâ’ is an example of these. if this splitting in the d orbitals is great enough electrons will fill low lying orbitals, pairing with other electrons in a given orbital, before filling higher energy orbitals. in question 7 we had fe2+, furthermore we found that there were a certain number (non-zero) of unpaired electrons. consider now fe(cn)6 4â’: here we also have fe2+, but in this case all the electrons are paired, yielding a diamagnetic species. how can you explain this?