Question

The steering rockets in the Space Shuttle use \(\mathrm{N}_{2} \mathrm{O}_{4}\) and a derivative of hydrazine, 1,1 -dimethylhydrazine (page 245 ). This mixture is called a hypergolic fuel because it ignites when the reactants come into contact: $$\begin{aligned} \mathrm{H}_{2} \mathrm{NN}\left(\mathrm{CH}_{3}\right)_{2}(\ell)+2 \mathrm{N}_{2} \mathrm{O}_{4}(\ell) & \rightarrow \\ 3 \mathrm{N}_{2}(\mathrm{g}) &+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$ (a) Identify the oxidizing agent and the reducing agent in this reaction. (b) The same propulsion system was used by the Lunar Lander on moon missions in the 1970 s. If the Lander used \(4100 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{NN}\left(\mathrm{CH}_{3}\right)_{2},\) what mass (in kilograms) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) was required to react with it? What mass (in kilograms) of each of the reaction products was generated?

Short answer

(a) Oxidizing: \(\mathrm{N}_2\mathrm{O}_4\), Reducing: \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\). (b) \(12556.60\;\mathrm{kg}\) of \(\mathrm{N}_2\mathrm{O}_4\), products: \(5731.83\;\mathrm{kg}\;\mathrm{N}_2\), \(4916.43\;\mathrm{kg}\;\mathrm{H}_2\mathrm{O}\), \(6005.99\;\mathrm{kg}\;\mathrm{CO}_2\).

Step-by-step solution

Identify the Oxidizing and Reducing Agents

To identify the oxidizing and reducing agents, we first need to understand oxidation-reduction (redox) reactions, where one substance loses electrons (is oxidized) and another gains electrons (is reduced). Here, hydrazine derivative \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\) is oxidized (loses hydrogen, gains oxygen), and \(\mathrm{N}_2\mathrm{O}_4\) is reduced (loses oxygen molecules, increases nitrogen). Thus, \(\mathrm{N}_2\mathrm{O}_4\) acts as the oxidizing agent, and \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\) acts as the reducing agent.

Determine Initial Masses and Moles for Reaction

Convert the given mass of \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\) into moles to find how much \(\mathrm{N}_2\mathrm{O}_4\) is needed. The molar mass of \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\) is calculated as follows: \((2\times1.01) + (2\times14.01) + (2\times12.01) + (8\times1.01) = 60.1\;\mathrm{g/mol}\). For \(4100\;\mathrm{kg}\), which is \(4100000\;\mathrm{g}\), the moles are \(\frac{4100000}{60.1} = 68219.64\;\text{moles}\).

Calculate corresponding mass of \(\mathrm{N}_2\mathrm{O}_4\)

Using the reaction stoichiometry, for each mole of \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\), 2 moles of \(\mathrm{N}_2\mathrm{O}_4\) are required. This means we need \(2 \times 68219.64 = 136439.28\;\text{moles of}\;\mathrm{N}_2\mathrm{O}_4\). The molar mass of \(\mathrm{N}_2\mathrm{O}_4\) is \((2\times14.01) + (4\times16.00) = 92.02\;\mathrm{g/mol}\). The total mass is \(136439.28\times92.02 = 12556598.3\;\mathrm{g}\;\approx 12556.60\;\mathrm{kg}\).

Calculate the Mass of Each Reaction Product

Given the reaction products \(3 \mathrm{N}_2\), \(4 \mathrm{H}_2\mathrm{O}\), and \(2 \mathrm{CO}_2\), we find their masses likewise using stoichiometry. **Nitrogen (\(\mathrm{N}_2\))** For \(3 \mathrm{N}_2\) from each \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\), we have \(3\times68219.64 = 204658.92\;\text{moles of}\;\mathrm{N}_2\). Molar mass of \(\mathrm{N}_2\) is \(28.02\;\mathrm{g/mol}\), hence the mass is \(204658.92\times28.02 = 5731834.7\;\mathrm{g} = 5731.83 \;\text{kg}\).**Water (\(\mathrm{H}_2\mathrm{O}\))** For \(4 \mathrm{H}_2\mathrm{O}\), we have \(4\times68219.64 = 272878.56\;\text{moles}\). Molar mass of \(\mathrm{H}_2\mathrm{O}\) is \(18.02\;\mathrm{g/mol}\), giving a mass of \(272878.56\times18.02 = 4916425.9\;\mathrm{g} = 4916.43\;\text{kg}\).**Carbon Dioxide (\(\mathrm{CO}_2\))** \(2\mathrm{CO}_2\) gives us \(2\times68219.64 = 136439.28\;\text{moles}\). Molar mass of \(\mathrm{CO}_2\) is \(44.01\;\mathrm{g/mol}\), resulting in a mass of \(136439.28\times44.01 = 6005988.0\;\mathrm{g} = 6005.99\;\text{kg}\).

Key concepts

Hypergolic Fuels

Hypergolic fuels are a fascinating aspect of rocket science. They are unique due to their ability to ignite spontaneously upon contact with an oxidizer. This property eliminates the need for an external ignition source, making them very reliable for space missions where robustness is crucial. In the case of the Space Shuttle, the hypergolic fuel used was a mixture of nitrogen tetroxide (\(\mathrm{N}_2\mathrm{O}_4\)) and 1,1-dimethylhydrazine.
These types of fuels have key advantages:

• Instant ignition, reducing system complexity.
• Prolonged shelf life and stability at room temperature.
• Usability in space, where traditional ignition sources may fail.

However, they are also highly toxic and require careful handling to ensure safety during storage and usage. Understanding how hypergolic fuels function is critical for engineering reliable propulsion systems in aerospace technology.

Oxidizing and Reducing Agents

In redox reactions, identifying oxidizing and reducing agents is essential. These agents drive the electron transfer process that characterizes these reactions. In the given equation, two key players are involved: 1,1-dimethylhydrazine and nitrogen tetroxide.
The oxidizing agent is the substance that gains electrons and is reduced in the process. In this reaction, \(\mathrm{N}_2\mathrm{O}_4\) is the oxidizing agent because it gains electrons by losing oxygen molecules. On the other hand, the reducing agent loses electrons and is oxidized. Here, \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\) serves as the reducing agent, donating electrons as it interacts with the oxidizer.
To summarize:

• Oxidizing Agent: \(\mathrm{N}_2\mathrm{O}_4\)
  
• Reducing Agent: 1,1-dimethylhydrazine

Understanding these roles helps in balancing redox equations and predicting the outcomes of chemical reactions.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. This is crucial for predicting the amounts required and produced in a given reaction. In stoichiometry, balanced chemical equations are used to derive the mole ratios needed for calculations.
In this example, the reaction equation shows that 1 mole of 1,1-dimethylhydrazine reacts with 2 moles of nitrogen tetroxide to produce products. This ratio helps to determine the amount of reactants needed.

• From the given equation: 1 \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\): 2 \(\mathrm{N}_2\mathrm{O}_4\)

By understanding the stoichiometry, we can calculate the masses of reactants required and predict the masses of products generated, ensuring efficient resource use in chemical and industrial processes.

Molar Mass Calculations

Molar mass calculations are essential for converting between the mass of a substance and the number of moles, which is crucial for stoichiometric equations. The molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol).
To perform these calculations accurately:

• Calculate the molar mass of each compound by summing the atomic masses of all the atoms in its formula.
• For 1,1-dimethylhydrazine (\(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\)), add the atomic masses to get 60.1 g/mol.
• For nitrogen tetroxide (\(\mathrm{N}_2\mathrm{O}_4\)), adding the respective atomic masses results in 92.02 g/mol.

These molar mass values then allow you to convert the mass of substances in grams to moles and vice versa. This conversion helps determine how much of each reactant is needed and predicts the yield of products in chemical reactions.