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(A) $1$. Calculate the millimoles of $H_2SO_4$ taken: $50 \ mL 	imes 0.5 \ M = 25 \ mmol$. $2$. Calculate the millimoles of $NaOH$ used to neutralize

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$500$

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(A) $1$. Calculate the millimoles of $H_2SO_4$ taken: $50 \ mL 	imes 0.5 \ M = 25 \ mmol$. $2$. Calculate the millimoles of $NaOH$ used to neutralize the excess $H_2SO_4$: $60 \ mL 	imes 0.5 \ M = 30 \ mmol$. $3$. Since $2 \ mol$ of $NaOH$ neutralize $1 \ mol$ of $H_2SO_4$,the $H_2SO_4$ neutralized by $NaOH$ is $30 / 2 = 15 \ mmol$. $4$. Millimoles of $H_2SO_4$ reacted with ammonia = $25 - 15 = 10 \ mmol$. $5$. Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,the millimoles of $NH_3$ (and thus $N$) = $10 	imes 2 = 20 \ mmol$. $6$. Mass of nitrogen = $20 \ mmol 	imes 14 \ g/mol = 280 \ mg$. $7$. Percentage of nitrogen = $(	ext{Mass of } N / x) 	imes 100 = 56$. $8$. $(280 / x) 	imes 100 = 56 \implies x = 28000 / 56 = 500 \ mg$.

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