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 equations for the process is `4NH_(3)(g) + 5O_(2)(g) rarr 4NO(g). + 6H_(2)O (g) 17 xx 4 = 68g 32 xx 5= 160g 30 xx 4= 120g`
According to the equation, 68g of `NH_(3)` require 160g of `O_(2)` for oxidation. Therefore, 10g of `NH_(3)` would required `= (160)/(68 ) xx 10` `=23.53 g " of " O_(2)`
SInce, the used oxygen is only 20g, oxygen is the limiting agent. Because, 160g of `O_(2)` gives NO= 120g
Thereofre, 20.00 g of `O_(2)` give NO `= (120)/(160) xx 20= 1.500g`