What happen when $N a O H$ reacts with?
$N H_{4} C l$
$N H_{4} N O_{3}$
$(N H_{4})_{2} S O_{2}$
Urea

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Solution

(i) Ammonium chloride reacts with $N a O H$ to produce $N H_{3}$

$N H_{4} C l + N a O H ⟶ N H_{3} + N a C l + H_{2} O$

(ii) Ammonium nitrate reacts with $N a O H$ to produce $N H_{3}$

$N H_{4} N O_{3} + N a O H ⟶ N H_{3} + N a N O_{3} + H_{2} O$

(iii) Ammonium sulphate also produces $N H_{3}$ with $N a O H$

$(N H_{4})_{2} S O_{4} + N a O H ⟶ N a_{2} S O_{4} + N H_{3} + H_{2} O$

(iv) Urea also produces $N H_{3}$ with $N a O H$ .

$H_{2} N - O | | C - N H_{2} Urea {O H}^{-} - -- \to O H - O | | C - N H_{2} + N H_{3}$

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