| Nitric acid can be produced from \[N{{H}_{3}}\] in three step process |
| I. \[4N{{H}_{3}}(g)+5{{O}_{2}}(g)\xrightarrow{{}}4NO(g)+6{{H}_{2}}O(g)\] |
| II. \[2NO(g)+{{O}_{2}}(g)\xrightarrow{{}}2N{{O}_{2}}(g)\] |
| III. \[3N{{O}_{2}}(g)+{{H}_{2}}O(l)\xrightarrow{{}}2HN{{O}_{3}}(aq)+NO(g)\] |
| % yield of Ist, IInd and IIIrd are respectively 40%, 50% and 70% respectively, then what volume of\[N{{H}_{3}}(g)\] sit 1 atm and \[0{}^\circ C\] required to produce 1075 g of \[HN{{O}_{3}}\]? |
A) 3413 L
B) 3500 L
C) 6826 L
D) 1750 L
Correct Answer: A
Solution :
| [a] Moles of \[N{{O}_{2}}\] required |
| \[=\left( \frac{1075}{63} \right)\times \frac{3}{2}\times \frac{1}{0.7}=35.56g\] |
| Moles of NO required\[=\frac{36.56}{0.6}\] |
| Moles of \[N{{N}_{3}}\] required |
| \[=\frac{36.56}{0.6}\times \frac{1}{0.40}=152.325\] |
| Volume of \[N{{H}_{3}}\] at STP required |
| \[=152.325\times 22.4\] |
| \[\simeq 3413L\] |
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