A) 22.4 KOH
B) 1.01 KOH
C) 6.02 KOH
D) 11.2 KOH
Correct Answer: D
Solution :
| \[HN{{O}_{3}}+KOH\to KN{{O}_{3}}+{{H}_{2}}O\] |
| \[\frac{12.6}{63}=\]0.2 mole; \[HN{{O}_{3}}\equiv KOH\] |
| 0.2 mole \[\equiv \] 0.2 mole |
| \[0.2\times 56=11.2gm\]. |
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