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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?

Respuesta :

barnuts

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.
Oseni

The limiting reagent would be O2

Limiting reagents

They are reagents that determine how much of the product is formed from a reaction.

From the balanced equation of the reaction:

4NH3 + 5O2 → 4NO + 6H2O

The mole ratio of NH3 to O2 is 4:5.

mole of 40.0 g NH3 = 40/17

                                     = 2.35 mole

Mole of 50.0 g O2 = 50/32

                               = 1.56 moles

Thus, NH3 seems to be in excess, making O2 to be the limiting reagent.

More on limiting reagents can be found here: https://brainly.com/question/11848702

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