How many liters of NH₃, at STP, will react with 8.5 g O₂ to form NO₂ and water?

Answer 1

4NH₃ + 7O_2. ==> 4NO₂ + 6H₂O balanced equation

moles O₂ used = 8.5 g O₂ x 1 mole/32 g = 0.266 moles O₂

moles NH₃ needed = 0.266 mol O₂ x 4 mol NH₃/7 mol O₂ = 0.152 moles

At STP, 0.152 moles x 22.4 L/mole = 3.4 liters of NH₃

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