Question

Under the right condition ammonia can be converted to nitrogen monoxide NO, NH3 + O2 =====> NO + H2O
HOW MANY GRAM OF H2OCOULD BE FORMED FROM A MIXTURE OF 51g OF NH3 AND 48g OF O2 ?

Answer 1

Given : Under the right condition ammonia can be converted into nitrogen monoxide.

i.e., 4NH₃ + 5O₂ ⇒4NO + 6H₂O

To find : amount of hydrogen could be formed from a mixture of 51g of NH₃ and 48g of O₂.

solution : given, weight of Ammonia = 51g

so, number of moles of ammonia = 51/17 = 3

weight of oxygen = 48g

so, number of moles of oxygen = 48/32 = 1.5 moles.

balanced chemical reaction is ...

4NH₃ + 5O₂ ⇒4NO + 6H₂O

here it is clear that 4 moles of ammonia react with 5 moles of oxygen and form 6 moles of water.

so, 3 moles of ammonia react with 5/3 moles of oxygen.

so oxygen is limiting reagent.

Therefore reaction based on oxygen.

here 5 moles of oxygen form 6 moles of water.

so, 1.5 moles of oxygen form 6/5 = 1.2 moles of water.

therefore weight of water = 1.2 × 18 = 21.6 g

Answer 2

we know that 68g of NH3 needs 160g of O2

HENCE,51g of NH3 needs 120g of O2 ,but we have only 48g which is less than requrement,hence O2 is the limiting reagent.