For a reaction, N 2 g + 3 H 2 ( g ) ⟶ 2 N H 3 ( g ) , identify di-hydrogen ( H 2 ) as a limiting reagent in the following reaction mixtures.

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N 2 g + 3 H 2 g → 2 N H 3 g moles of N 2 stoichiometry of N 2 ≠ mole of H 2 stoichiometry of H 2 Whose ratio is less called limiting reagent. (a) mole of N 2 = 28 28 = 1 m o l e s s t i o c h i o m e t r y ratio of N 2 = 1 1 = 1 mole of H 2 = 6 2 = 3 m o l e s s t i o c h i o m e t r y ratio of H 2 = 3 3 = 1 there is no limiting reagent. (b) mole of N 2 = 35 28 = 1.25 m o l e s s t i o c h i o m e t r y ratio of N 2 = 1.25 1 limiting reagent mole of H 2 = 8 2 = 4 m o l e s s t i o c h i o m e t r y ratio of H 2 = 4 3 = 1.33 (c) mole of N 2 = 56 28 = 2 m o l e s s t i o c h i o m e t r y ratio of N 2 = 2 1 = 2 mole of H 2 = 10 2 = 5 m o l e s s t i o c h i o m e t r y ratio of H 2 = 5 3 = 1.66 limiting reagent. (d) mole of N 2 = 14 28 = 0.5 m o l e s s t i o c h i o m e t r y ratio of N 2 = 0.5 1 = 0.5 limiting reagent mole of H 2 = 4 2 = 2 m o l e s s t i o c h i o m e t r y ratio of H 2 = 2 3 = 1.66 ∴ Answer is 3 .

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