In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia 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` g of oxygen ?

Explanation

The balanced chemical equation for the process is
`underset(17xx4=68g)(4NH_(3)(g))+underset(32xx5=160g)(5O_(2)(g))tounderset(30xx4=120g)(4NO(g))+6H_(2)O(g)`
According to the equation, 68 g of NH3 require 160 g of `O_(2)` for oxidation. Therefore, 10 g of `NH_(3)` would require `(160)/(68)xx10=23.53g` of `O_(2)`. Since, the used `O_(2)` is only 20 g. `O_(2)` is the limiting reagent.
`:.160g` of `O_(2)` gives N=120g
`:.20.00g` of `O_(2)` will give NO `=(120)/(160)xx20.00`
15.00g.