In Ostwaid's process for the manufacture of nitric acid , the first step involves the oxidation of anmonia gas by oxygen gas to give nitric oxide gas and steam . What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 if oxygen ?

Explanation

Reaction : `underset(4xx17g=68g)(4NH_(3)(g))+underset(5xx32g=160g)(5O_(2)(g))tounderset(4xx30g=120)(4NO(g))+6H_(2)O(g)`
Therefore n160g `O_(2)` is required to oxidised 68g `NH_(3)`
`therefore`,20`O_(2)`is required to oxidise `(66)/(100)xx20`g of `NH_(3)`
=8.5g of `NH_(3)`
So here `O_(2)` is th limiting reagent . Amount of nitric oxide produced depends upon the amount of oxygen taken and not on the amount of `NH_(3)`taken.
`therefore` According to the above equation
160 g `O_(2)`produce 120 g NO
`therefore`20g `O_(2)` produces `(120)/(160)xx20 g` NO=15g NO
Thus the reaction between 10g `NH_(3)` and 20g `O_(2)` produces a maximum amount of 15 g NO.