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(B) The reaction between $NH_{3}$ and $H_{2}SO_{4}$ is: $2NH_{3} + H_{2}SO_{4} \rightarrow (NH_{4})_{2}SO_{4}$.$1 \ mol$ of $H_{2}SO_{4}$ neutralizes

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(B) The reaction between $NH_{3}$ and $H_{2}SO_{4}$ is: $2NH_{3} + H_{2}SO_{4} ightarrow (NH_{4})_{2}SO_{4}$.$1 \ mol$ of $H_{2}SO_{4}$ neutralizes $2 \ mol$ of $NH_{3}$.$10 \ mL$ of $1 \ M$ $H_{2}SO_{4}$ contains $10 \ mmol$ of $H_{2}SO_{4}$.Therefore,$NH_{3}$ produced $= 2 	imes 10 \ mmol = 20 \ mmol = 0.02 \ mol$.Mass of nitrogen $= 0.02 \ mol 	imes 14 \ g/mol = 0.28 \ g$.Percentage of nitrogen $= \frac{	ext{Mass of Nitrogen}}{	ext{Mass of Compound}} 	imes 100 = \frac{0.28}{0.5} 	imes 100 = 56 \ \%$.

During the estimation of nitrogen present in an organic compound by the Kjeldahl method,the ammonia evolved from $0.5 \ g$ of the compound neutralized $10 \ mL$ of $1 \ M$ $H_{2}SO_{4}$. Find the percentage of nitrogen in the compound. (in $\%$)

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