MCQ
How many .................. $\mathrm{gm}$ $\mathrm{KOH}$ grams of caustic potash required to completely neutralise $12.6\, gm$ $HN{O_3}$
- A$22.4 $
- B$1.01$
- C$6.02$
- ✓$11.2$
✓
Answer
Correct option: D.
$11.2$
d
(d) $HN{O_3} + KOH \to KN{O_3} + {H_2}O$
(d) $HN{O_3} + KOH \to KN{O_3} + {H_2}O$
$\frac{{12.6}}{{63}} = 0.2$ mole; $HN{O_3} \equiv KOH$
$0.2$ mole $ \equiv $ $0.2$ mole
$0.2 \times 56 = 11.2\,gm$.
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