Question

consider the Haber process through which fertilizer is made:

              N₂(g)                +     3 H₂(g)            Þ               2 NH₃

            f.w. 28.0                    f.w.2.0                           f.w. 17.0

If you have 32.0 g of N₂ and 8.0 g of H₂, which will be the limiting reagent? Support your answer with appropriate calculations.

Answer 1

* Haber's Process Na g) + 31.10) ► NHs (9) formula meight of Ne weight of 28.0 2.0 formula Mass Mass of N2 H₂ 8.og j of Number moles of Ginen mass. Formula meight: 32 28 1. 14285 mole Nz . Number of moles of Hi Given Moss Formula weight 8 2 4 moles H₂

o even Linniting Reagent is the substance at reactant Because limiting which limits or stops the reaction. heagent 'greets completely other species cet reactant side rumains in reacted The compound ). species reactant side which is auailable less than the Calculated value is the limiting roagent. Y Now from the about balanced cheuscal I realtion clear that Mole Dreacts Twith وزن of 3 moles of H₂. 1 mole Na reacts with 3 moles of H₂ 1.14285 mole Ni will react with = 38 1.14285 mole H₂ = 3.42.855 mole H₂ * But have 4 moles which is more than the calculated value Th1Jort hau H2 is PXCESS. rence, Ng is the limiting lieagent. can also prove as follows: 3 moles H₂ reacts with I mole of Ne mole H₂ reacts ичи mole श्री 14 moles of H₂ geacts with a vole 1.3333 mole N₂

- But we have is less than the only 1.14285 mole of Ny which required calculated value Therefore, Ng is the limiting Reagent

Therefore, N_{​​​​​​2} will be the limiting reagent.

Thank You.