A 58.3g sample of NH3 is reacted with 126g O2, according to this reaction what is the limiting reagent? 4NH3 + 7O2 --> 4NO + 6H2O

Explanation:

4NH₃      +     7O₂     -->     4NO   +     6H₂O

4 moles       7 moles

58.3 g of NH₃ = 58.3 / 17 = 3.43 moles

126 g of O₂ = 126 / 32 = 3.94 moles

4 moles NH₃ requires 7 moles of O₂

3.43 moles NH₃ requires 6 moles O₂

Available O₂ is only 3 94 moles , so O₂ is in short supply .

Hence O₂ is the limiting reagent .

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