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 balance equation for the reaction is:
`4NH_(3)(g)+5O_(2)(g)rarr 4NO(g)+6H_(2)O(g)`
`{:(4xx17,,5xx32,,4xx30),(=68 g,,=160 g,,=120 g):}`
Here, `160 g` of `O_(2)` will react with `NH_(3)=68 g`
`:. 20 g` of `O_(2)` will react with `NH_(3)=68/160xx20=8.5 g`
Thus, `O_(2)` is the limiting reagent. Therefore, the calculate must be based upon the amount of `O_(2)` taken and not on the amount of `NH_(3)` taken.
from the equation,
`160 g` of `O_(2)` produce `NO=120 g`
`:. 20 g` of `O_(2)` will produce `NO` will produce `NO=120/160xx20=15 g`