Question

The steering rockets in space vehicles use \(\mathrm{N}_{2} \mathrm{O}_{4}\) and a derivative of hydrazine, 1,1 -dimethylhydrazine (Study Question \(5.86) .\) 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

Oxidizing agent: \(\mathrm{N}_2 \mathrm{O}_4\), Reducing agent: \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\). Required \(\mathrm{N}_2 \mathrm{O}_4\): 12550 kg. Products: 5732.68 kg \(\mathrm{N}_2\), 4914.42 kg \(\mathrm{H}_2 \mathrm{O}\), 6004.82 kg \(\mathrm{CO}_2\).

Step-by-step solution

Identify the Oxidizing and Reducing Agents

In a chemical reaction, the oxidizing agent is the substance being reduced and gains electrons, while the reducing agent is the substance being oxidized and loses electrons. Here, nitrogen dioxide (\(\mathrm{N}_2\mathrm{O}_4\)) gains electrons and is reduced to \(\mathrm{N}_2\), thus acting as the oxidizing agent. \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\) is oxidized to form \(\mathrm{N}_2\) and other products, making it the reducing agent.

Write the Balanced Chemical Equation

The balanced equation is given as: \[ \mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2(\ell) + 2 \mathrm{N}_2 \mathrm{O}_4(\ell) \rightarrow 3 \mathrm{N}_2(\mathrm{g}) + 4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) + 2 \mathrm{CO}_2(\mathrm{g}) \] This equation tells us that 1 mole of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) reacts with 2 moles of \(\mathrm{N}_2 \mathrm{O}_4\) to produce the products.

Calculate Moles of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\)

First, find the molar mass of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\): \[(2 \cdot 1.01) + (2 \cdot 14.01) + (2 \cdot 12.01) + (6 \cdot 1.01) = 60.1 \, \text{g/mol}\] For 4100 kg which is 4100000 g: \[\frac{4100000 \, \text{g}}{60.1 \, \text{g/mol}} = 68220 \text{ moles}\]

Calculate Moles of \(\mathrm{N}_2 \mathrm{O}_4\) Required

From the balanced equation, 1 mole of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) reacts with 2 moles of \(\mathrm{N}_2 \mathrm{O}_4\). So, \[68220 \, \text{moles of } \mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2 \times 2 = 136440 \, \text{moles of } \mathrm{N}_2 \mathrm{O}_4\]

Calculate Mass of \(\mathrm{N}_2 \mathrm{O}_4\) Required

Find the molar mass of \(\mathrm{N}_2 \mathrm{O}_4\): \[(2 \cdot 14.01) + (4 \cdot 16.00) = 92.02 \, \text{g/mol}\] \[136440 \, \text{moles} \times 92.02 \, \text{g/mol} = 12550 imes 10^3 \, \text{g} = 12550 \, \text{kg}\]

Calculate Mass of \(\mathrm{N}_2\) Produced

From the balanced equation, 1 mole of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) produces 3 moles of \(\mathrm{N}_2\). \[68220 \, \text{moles} \times 3 = 204660 \, \text{moles of } \mathrm{N}_2\] Molar mass of \(\mathrm{N}_2\) is \(28.02 \, \text{g/mol}\). \[204660 \, \text{moles} \times 28.02 \, \text{g/mol} = 5732.68 \times 10^3 \, \text{g} = 5732.68 \, \text{kg}\]

Calculate Mass of \(\mathrm{H}_2 \mathrm{O}\) Produced

From the equation, 1 mole of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) produces 4 moles of \(\mathrm{H}_2 \mathrm{O}\). \[68220 \, \text{moles} \times 4 = 272880 \, \text{moles of } \mathrm{H}_2 \mathrm{O}\] Molar mass of \(\mathrm{H}_2 \mathrm{O}\) is \(18.02 \, \text{g/mol}\). \[272880 \, \text{moles} \times 18.02 \, \text{g/mol} = 4914.42 \times 10^3 \, \text{g} = 4914.42 \, \text{kg} \]

Calculate Mass of \(\mathrm{CO}_2\) Produced

From the equation, 1 mole of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) produces 2 moles of \(\mathrm{CO}_2\). \[68220 \, \text{moles} \times 2 = 136440 \, \text{moles of } \mathrm{CO}_2\] Molar mass of \(\mathrm{CO}_2\) is \(44.01 \, \text{g/mol}\). \[136440 \, \text{moles} \times 44.01 \, \text{g/mol} = 6004.82 \times 10^3 \, \text{g} = 6004.82 \, \text{kg} \]

Key concepts

Oxidizing and Reducing Agents

In chemical reactions, there are substances called oxidizing and reducing agents. These play crucial roles in the chemistry of substances, especially in energetic reactions like those involving hypergolic fuels. Let's explore how these agents work.

The oxidizing agent in a reaction gains electrons. This means it is reduced during the chemical process. In our example, nitrogen dioxide (\(\mathrm{N}_2\mathrm{O}_4\)) acts as the oxidizing agent. It accepts electrons and gets converted into \(\mathrm{N}_2\), a different form of nitrogen that makes up most of Earth's atmosphere.On the flip side, the reducing agent loses electrons and is oxidized. Here, \(\mathrm{H}_2\mathrm{NN}(\mathrm{CH}_3)_2\), a derivative of hydrazine, plays this role. It loses electrons while transforming into products like \(\mathrm{N}_2\), \(\mathrm{H}_2\mathrm{O}\), and \(\mathrm{CO}_2\). This dance of electrons between oxidizing and reducing agents is the heart of the redox (reduction-oxidation) reactions.

Understanding this electron exchange helps in identifying how reactions will proceed and which substances will form as products.

Chemical Reaction Balancing

Balancing chemical equations is a fundamental skill in chemistry. It ensures that the same number of each type of atom appears on both sides of the equation, which aligns with the law of conservation of mass. Let's see why this is crucial.

Balance is necessary because atoms are not created or destroyed in chemical reactions. Instead, they are rearranged. Consider our reaction of hypergolic fuel. When 1 mole of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) reacts with 2 moles of \(\mathrm{N}_2\mathrm{O}_4\), it produces exactly 3 moles of nitrogen gas \(\mathrm{N}_2\), 4 moles of water \(\mathrm{H}_2 \mathrm{O}\), and 2 moles of carbon dioxide \(\mathrm{CO}_2\).

Let’s put it simply: to balance equations, ensure each element has an equal number of atoms on both sides. This way, the chemical equation accurately reflects the conservation of mass, ensuring it provides correct stoichiometric relationships, vital for further calculations.

Mass Calculation in Chemical Reactions

When dealing with chemical reactions, especially in real-world applications like space flight, calculating mass accurately is essential. This involves understanding molar masses and stoichiometric relationships.

First, we calculate the amount of a substance using its molar mass – the weight of one mole of a chemical substance. For instance, the molar mass of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\) is 60.1 g/mol, and \(\mathrm{N}_2 \mathrm{O}_4\) is 92.02 g/mol. With a known mass of a starter substance, say 4100 kg of \(\mathrm{H}_2 \mathrm{NN}(\mathrm{CH}_3)_2\), we convert this to moles by dividing by its molar mass.

Next, stoichiometric coefficients from the balanced equation are used. For every mole of hydrazine derivative, 2 moles of \(\mathrm{N}_2 \mathrm{O}_4\) are required. Using this ratio helps calculate 136440 moles of \(\mathrm{N}_2 \mathrm{O}_4\), which corresponds to a specific mass.

Finally, each reaction product's mass is found similarly by using their molar masses and the stoichiometric relations. This completes the picture of how much of each substance is involved and produced, vital for logistical and scientific applications in chemical engineering and space missions.