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.]
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