Question

How many grams of caustic potash required to completely neutralise 12.6 gm $HN{O}_3$

• A. 22.4 KOH
• B. 1.01 KOH
• C. 6.02 KOH
• D. 11.2 KOH

Answer 1

The correct option is D 11.2 KOH

$HN{O}_3+KOH\to KN{O}_3+H_{2}O$
$\frac{12.6}{63}=0.2mole;HN{O}_3\equiv KOH$
$0.2mole\equiv 0.2mole$
$0.2\times 56=11.2gm$