Question

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.

Short answer

The mass percent of N2H4 in the original gaseous mixture is approximately 28.98%.

Step-by-step solution

Write the balanced chemical equations

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

Determine the initial number of moles of O2 consumed

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

Calculate the moles of NH3 and N2H4

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

Calculate the mass of NH3 and 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

Determine the mass percent of N2H4

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%.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is vital in understanding chemical reactions. It reflects the conservation of mass, meaning what goes into the reaction must come out. For instance, in the exercise, we are dealing with two reactions:

• The reaction between ammonia (\( \mathrm{NH}_3(g) \)) and oxygen (\( \mathrm{O}_2(g) \)) forms nitrogen dioxide (\( \mathrm{NO}_2(g) \)) and water (\( \mathrm{H}_2 \mathrm{O}(g) \)). The balanced equation is 4NH\(_3\) + 7O\(_2\) \( ightarrow\) 4NO\(_2\) + 6H\(_2\)O.
• Hydrazine (\( \mathrm{N}_2\mathrm{H}_4(g) \)) reacting with oxygen (\( \mathrm{O}_2(g) \)) also forms \( \mathrm{NO}_2(g) \) and \( \mathrm{H}_2 \mathrm{O}(g) \), with the equation: 2N\(_2\)H\(_4\) + 7O\(_2\) \( ightarrow\) 4NO\(_2\) + 4H\(_2\)O.

Each equation shows that for every set amount of reactants, a specific amount of products will form. Balancing equations ensure that each atom is accounted for, and no mass is lost in the process.Balanced equations help in calculating the exact amounts of substances involved, like how many moles of \( \mathrm{O}_2 \) are consumed by each reactant.

Mass Percent Calculation

Mass percent is a way of expressing a concentration of a component in a mixture or a compound. It's calculated by dividing the mass of the component by the total mass of the mixture, then multiplying by 100%.In the exercise, after determining the mass of \( \mathrm{N}_2 \mathrm{H}_4(g) \) that reacted, you can calculate its mass percent in the original mixture. We found the mass of \( \mathrm{N}_2\mathrm{H}_4 \) to be 17.68g:So, the mass percent calculation is:Mass percent = \( \left( \frac{\text{Mass of } \mathrm{N}_2\mathrm{H}_4}{\text{Total mass of mixture}} \right) \times 100\% \)Plug in the values:Mass percent = \( \left( \frac{17.68}{61.00} \right) \times 100\% \approx 28.98\% \) This shows that \( \mathrm{N}_2\mathrm{H}_4 \) constitutes approximately 28.98% of the original mixture's mass. Understanding mass percent is crucial because it gives a clear picture of the composition of mixtures, important for chemical formulations, labeling, and industrial processes.

Moles and Mass Relationship

The concept of moles is essential to connect the microscopic particles of a substance to its measurable mass. A mole is a unit that represents \(6.022 \times 10^{23} \) particles (Avogadro's number).In this exercise, you calculate the moles of each gas (whether \( \mathrm{NH}_3 \) or \( \mathrm{N}_2\mathrm{H}_4 \)) and relate it to their masses using molar mass — a conversion factor between mass (grams) and moles.Here's how it translates:

• For ammonia \( \mathrm{NH}_3 \): Molar mass = 17.03 g/mol. Calculated number of moles converted to mass: Mass = moles × molar mass = 2.074 × 17.03 = 35.32 g.
• For hydrazine \( \mathrm{N}_2\mathrm{H}_4 \):Molar mass = 32.05 g/mol. Similarly, converting moles to mass: Mass = moles × molar mass = 0.552 × 32.05 = 17.68 g.

This relationship allows for quantitative analysis in chemical equations by converting moles to measurable, tangible values.Understanding this relationship is key for scientists to measure and predict chemical behavior accurately in labs and industries.