Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4nh3 + 5o2 → 4no + 6h2o when 40.0 g nh3 and 50.0 g o2 are allowed to react, which is the limiting reagent?

Question

Hoffman

Chemistry

2 June, 10:13

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4nh3 + 5o2 → 4no + 6h2o when 40.0 g nh3 and 50.0 g o2 are allowed to react, which is the limiting reagent?

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Answers (1)

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Isis Kidd · Answer 1

Convert the mass to moles:

40.0 grams of NH3 / 17.03 g/mol = 2.35 moles NH3

50.0 grams of O2 / 31.998 g/mol = 1.56 moles O2

Get the actual ratio:

NH3/O2 = 2.35 / 1.56 = 1.5

The theoretical ratio is 4 NH3 to 5 O2, that is:

(NH3/O2) theo = 4 / 5 = 0.8

Since actual > theoretical and the ratio is NH3/O2, therefore the limiting reagent is O2.