Problem 51 - Chemistry and Chemical Reactivity

Question

A major use of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) is in steam boilers in power plants. (a) The reaction of hydrazine with \(\mathrm{O}_{2}\) dissolved in water gives \(\mathrm{N}_{2}\) and water. Write a balanced equation for this reaction. (b) \(\mathrm{O}_{2}\) dissolves in water to the extent of \(3.08 \mathrm{cm}^{3}\) (gas at STP) in \(100 .\) mL of water at \(20^{\circ} \mathrm{C} .\) To consume all of the dissolved \(\mathrm{O}_{2}\) in \(3.00 \times 10^{4}\) I. of water (enough to fill a small swimming pool), what mass of \(\mathrm{N}_{2} \mathrm{H}_{4}\) is needed?

Step-by-Step Solution

Answer

NH4 + O2 -> N2 + 2H2O; 1320.60 g N2H4 is needed.

Step 1: Write the Unbalanced Chemical Equation

The reaction given is between hydrazine \( \mathrm{N}_2\mathrm{H}_4 \) and oxygen \( \mathrm{O}_2 \) to produce nitrogen \( \mathrm{N}_2 \) and water \( \mathrm{H}_2\mathrm{O} \). The unbalanced equation is: \[ \mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + \mathrm{H}_2\mathrm{O} \]

Step 2: Balance the Chemical Equation

To balance, first consider the nitrogen atoms. Each side needs 2 nitrogen atoms, which there are already, so they are balanced. Next, balance the hydrogen atoms: there are 4 hydrogens in \( \mathrm{N}_2\mathrm{H}_4 \), so we need 2 water molecules (each contributes 2 hydrogens) on the product side. This gives us: \[ \mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O} \] Finally, balance the oxygen atoms: Since 2 water molecules need 2 oxygen atoms, we need 1 oxygen molecule, \( \mathrm{O}_2 \), for the left-hand side. The balanced equation is: \[ \mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O} \]

Step 3: Calculate the Volume of O2 Consumed

Given that \( 3.08 \mathrm{cm}^3 \) of \( \mathrm{O}_2 \) dissolves per \( 100 \mathrm{mL} \) of water, we need to find the total volume of \( \mathrm{O}_2 \) dissolved in \( 3.00 \times 10^4 \) L of water. First, convert liters to milliliters: \[ 3.00 \times 10^4 \mathrm{L} = 3.00 \times 10^7 \mathrm{mL} \] Now, find the total \( \mathrm{O}_2 \): \[ \frac{3.08 \mathrm{cm}^3}{100 \mathrm{mL}} \times 3.00 \times 10^7 \mathrm{mL} = 9.24 \times 10^5 \mathrm{cm}^3 \]

Step 4: Convert Volume to Moles of O2

Use the molar volume of gas at STP, which is \( 22.414 \mathrm{L/mol} \), to convert \( \mathrm{O}_2 \) volume to moles. Convert \( \mathrm{cm}^3 \) to \( \mathrm{L} \): \[ 9.24 \times 10^5 \mathrm{cm}^3 = 9.24 \times 10^2 \mathrm{L} \] Calculate moles: \[ \frac{9.24 \times 10^2 \mathrm{L}}{22.414 \mathrm{L/mol}} \approx 41.22 \mathrm{mol} \ \mathrm{O}_2 \]

Step 5: Calculate Moles of N2H4 Required

According to the balanced chemical equation, 1 mol of \( \mathrm{O}_2 \) reacts with 1 mol of \( \mathrm{N}_2\mathrm{H}_4 \). Therefore, we need: \[ 41.22 \mathrm{mol} \ \mathrm{N}_2\mathrm{H}_4 \]

Step 6: Convert Moles of N2H4 to Mass

Find the molar mass of \( \mathrm{N}_2\mathrm{H}_4 \) using the atomic masses: each nitrogen is \( 14.01 \mathrm{g/mol} \) and each hydrogen is \( 1.01 \mathrm{g/mol} \). Total is: \[ 2(14.01) + 4(1.01) = 32.05 \mathrm{g/mol} \] Calculate the mass required: \[ 41.22 \mathrm{mol} \times 32.05 \mathrm{g/mol} \approx 1320.60 \mathrm{g} \]

Step 7: Final Answer

To consume all the dissolved \( \mathrm{O}_2 \), \( 1320.60 \mathrm{g} \) of \( \mathrm{N}_2\mathrm{H}_4 \) is needed.

Key Concepts

Balancing Equations

Chemical reactions, at their core, describe how substances transform into new substances. Balancing equations is crucial because it ensures the conservation of mass; in other words, atoms are neither created nor destroyed.
In our reaction, hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) reacts with oxygen (\(\mathrm{O}_2\)) to form nitrogen gas (\(\mathrm{N}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). Initially, the equation might not balance, presenting different numbers of atoms on each side.

• First, ensure each type of atom on the left matches those on the right.
• Adjust coefficients (the numbers in front of molecules, not the subscripts) to balance the atoms.
• Remember: Coefficients change the quantity of the whole molecule, not individual atoms.

For example, the unbalanced reaction:\[\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + \mathrm{H}_2\mathrm{O}\]
Adjust the water molecules to have 2, balancing hydrogen. Every new arrangement alters other atoms that must be adjusted as well, until you reach:\[\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O}\]
This balanced equation respects the Law of Conservation of Mass.

Stoichiometry

Stoichiometry is the heart of quantitative chemistry, linking reactants to products. Using the balanced chemical equation, we derive the relationship between quantities of reactants and products.
In our problem, knowing the balanced reaction...\[\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O}\]
...helps us determine how much hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) is required to react with a known amount of dissolved oxygen (\(\mathrm{O}_2\)). Here, 1 mole of \(\mathrm{N}_2\mathrm{H}_4\) reacts with 1 mole of \(\mathrm{O}_2\), simplifying calculations.
Steps include:

• Converting volumes of gases to moles using the molar volume.
• Using mole ratios from the balanced equation to relate reactants to each other.
• Converting moles of a substance into grams, by multiplying by its molar mass.

Thus, from the moles of oxygen we calculated, we found the equivalent moles and subsequently the mass of hydrazine necessary to complete the reaction.

Gas Laws

Gas laws help us understand how gases behave under various conditions.
One foundational concept, used often, is the molar volume of a gas at Standard Temperature and Pressure (STP). At STP, one mole of any gas occupies 22.414 liters.
This principle aids in converting between the volume of \(\mathrm{O}_2\) we have and the moles needed for stoichiometric calculations.
Key points of gas laws include:

• The relationship between pressure, volume, temperature, and number of moles (as described by the ideal gas law).
• Understanding standard conditions to simplify conversion calculations.

In our problem, we remained at standard conditions, so determining the moles of gas from volume was straightforward. This reflects the practical use of gas laws in ensuring precise and accurate chemical calculations.

Hydrazine Applications

Hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) is a versatile chemical with many important applications. Among its uses, it serves as a critical component in steam boilers for power plant operation.
This application is based on its chemical properties:

• It effectively scavenges oxygen dissolved in water, preventing corrosion of metal components in boilers.
• It reacts cleanly to form nitrogen gas (\(\mathrm{N}_2\)) and water, byproducts which are non-corrosive and inert, thus harmless in system operation.

The process helps maintain system efficiency and prolongs equipment lifespan by mitigating oxidative damage. Its properties, efficiency, and safety margin make hydrazine an ideal choice in these industrial settings. Recognizing these benefits highlights the broader context of why such reactions are indispensable beyond just theoretical chemical equations.