A gas contains a mixture of \(\mathrm{NH}_{3}(g)\) and $\mathrm{N}_{2} \mathrm{H}_{4}(g),\( both of which react with \)\mathrm{O}_{2}(g)$ to form \(\mathrm{NO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) The gaseous mixture (with an initial mass of 61.00 \(\mathrm{g}\) ) is reacted with 10.00 \(\mathrm{moles}\) \(\mathrm{O}_{2},\) and after the reaction is complete, 4.062 moles of \(\mathrm{O}_{2}\) remains. Calculate the mass percent of \(\mathrm{N}_{2} \mathrm{H}_{4}(g)\) in the original gaseous mixture.

For both reactions of NH3 and N2H4 with O2, we write down the balanced chemical equations: \( \mathrm{4NH}_3(g) + \mathrm{7O}_2(g) \rightarrow \mathrm{4NO}_2(g) + \mathrm{6H}_2\mathrm{O}(g) \) \( \mathrm{2N}_2\mathrm{H}_4(g) + \mathrm{7O}_2(g) \rightarrow \mathrm{4NO}_2(g) + \mathrm{4H}_2\mathrm{O}(g) \)

Initially, there were 10.00 moles of O2, and after the reactions, 4.062 moles of O2 remain. We can find the initial number of moles of O2 consumed by subtracting these values: Initial moles of O2 consumed = 10.00 moles - 4.062 moles = 5.938 moles

Let x moles of NH3 react and y moles of N2H4 react. According to the balanced chemical equations, For NH3: \(4x \: \mathrm{moles \: of \: NH}_3 \rightarrow 7x \: \mathrm{moles \: of \: O_2 \: consumed} \) For N2H4: \(2y \: \mathrm{moles \: of \: N_2H}_4 \rightarrow 7y \: \mathrm{moles \: of \: O_2 \: consumed} \) Add the two equations: \(7x + 7y = 4 + 3.938 = 5.938 \: \mathrm{moles \: of \: O_2(S_1)} \) We also know that the total mass of the gaseous mixture is 61.00 g, so using the molar masses of NH3 (\(17.03 \: \mathrm{g/mol}\)) and N2H4 (\(32.05 \: \mathrm{g/mol}\)): \(17.03x \: \mathrm{g} + 32.05y \: \mathrm{g} = 61.00 \: \mathrm{g \: (S_2)} \) To find the values of x and y, solve the system of equations: Solve \(S_1\) for y: \(y = \dfrac{5.938 - 7x}{7}\) Substitute y in \(S_2\) and solve for x: \(17.03x + 32.05(\dfrac{5.938 - 7x}{7}) = 61.00\) After solving for x, we get: x ≈ 2.074 moles of NH3 Substitute the value of x back into the expression for y: y ≈ 0.552 moles of N2H4

Using the molar masses and the number of moles for both compounds, we can calculate their masses: Mass of NH3 = 17.03 g/mol × 2.074 mol = 35.32 g Mass of N2H4 = 32.05 g/mol × 0.552 mol = 17.68 g

Now, we can find the mass percent of N2H4 in the initial gaseous mixture: Mass percent of N2H4 = (Mass of N2H4 / Total mass of mixture) × 100% Mass percent of N2H4 = (17.68 g / 61.00 g) × 100% ≈ 28.98% Thus, the mass percent of N2H4 in the original gaseous mixture is approximately 28.98%.