Calculation Steps:
Given Information:
Volume of \( \text{H}_2\text{SO}_4 \) = 10 mL = 0.01 L
Molarity of \( \text{H}_2\text{SO}_4 \) = 2M
Calculate moles of \( \text{H}_2\text{SO}_4 \):
\[\text{Moles of } \text{H}_2\text{SO}_4 = 2 \times 0.01 = 0.02 \, \text{mol}\]
Balanced Reaction:
\[2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4\]
Stoichiometry dictates that 2 moles of \( \text{NH}_3 \) react with 1 mole of \( \text{H}_2\text{SO}_4 \). Calculate moles of \( \text{NH}_3 \):
\[\text{Moles of } \text{NH}_3 = 2 \times 0.02 = 0.04 \, \text{mol}\]
Calculate mass of nitrogen in \( \text{NH}_3 \):
\[\text{Mass of nitrogen} = 0.04 \times 14 = 0.56 \, \text{g}\]
Calculate the percentage of nitrogen:
\[\text{Percentage of nitrogen} = \left( \frac{0.56}{1} \right) \times 100 = 56\%\]
Result: The compound contains \( 56\% \) nitrogen.