Question

In ostwald process for manufacture of nitric acid minimum weight of nitric oxide

Answer 1

                                       1100K

                 4NH₃ + 5O₂ --------------> 4NO + 6H₂O

                     |          |          Pt               |

             4 * 17g     5*32g                4*30g

             = 68g      = 160g               = 120g                       

 here you can see that,

     68g of NH₃ reacts with 160g of O₂

so, 10g of NH₃ reacts with 160/68 * 10 = 23.5 g of O₂

 

but here the available amount of O₂ is 20g, which is less than the required amount.hence, O₂ is limiting reagent.

 so, now,

  160g of O₂ produces 120g of  NO

so, 20g of O₂ produces 120/160 * 20 = 15g of NO

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