The ammonia (NH_{3}) released on quantitative | vedclass

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Question

(C) The chemical reaction is: $NH_{2}CONH_{2} + 2NaOH \rightarrow Na_{2}CO_{3} + 2NH_{3}$.$1 \ mole$ of urea produces $2 \ moles$ of $NH_{3}$.Molar ma

Vedclass · Accepted Answer

$100 \ mL$ of $0.2 \ N \ HCl$

Vedclass · Answer

(C) The chemical reaction is: $NH_{2}CONH_{2} + 2NaOH ightarrow Na_{2}CO_{3} + 2NH_{3}$.$1 \ mole$ of urea produces $2 \ moles$ of $NH_{3}$.Molar mass of urea $(NH_{2}CONH_{2}) = 14 + 2 + 12 + 16 + 14 + 2 = 60 \ g/mol$.Moles of urea $= \frac{0.6 \ g}{60 \ g/mol} = 0.01 \ mol$.Moles of $NH_{3}$ produced $= 2 	imes 0.01 = 0.02 \ mol$.For neutralization,moles of $HCl$ required $= 0.02 \ mol$.Checking options:Option $C$: $100 \ mL$ of $0.2 \ N \ HCl = 0.1 \ L 	imes 0.2 \ N = 0.02 \ 	ext{equivalents} = 0.02 \ 	ext{moles}$ of $HCl$.

The ammonia $(NH_{3})$ released on quantitative reaction of $0.6 \ g$ urea $(NH_{2}CONH_{2})$ with sodium hydroxide $(NaOH)$ can be neutralized by

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