question_answer 1)

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia by oxygen to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10 g ammonia and 20 g oxygen?

Answer:

[Hint: \[4\underset{68g}{\mathop{N{{H}_{3}}(g)}}\,+\underset{160g}{\mathop{5{{O}_{2}}(g)}}\,\to \underset{120g}{\mathop{4NO(g)}}\,+6{{H}_{2}}O(g)\] Case I: Let \[N{{H}_{3}}\] is completely consumed then mass of nitric oxide obtained will be: Mass of \[NO=\frac{120}{68}\times 10=\frac{1200}{68}g=17.64\,g\] Case II: Let \[{{O}_{2}}\] is completely consumed then mass of nitric oxide obtained will be: Mass of \[NO=\frac{120}{160}\times 20=15g\] Since, \[{{O}_{2}}\] gives least amount of product hence it is limiting and actual amount of NO will be 15 g.]