Question

Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}\right]\) mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)$$ If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(127^{\circ} \mathrm{C}\) in an evacuated \(250-\mathrm{L}\) tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has \(100 \%\) yield?

Short answer

The partial pressure of nitrogen gas produced is 9.89 atm, and the total pressure in the tank is 29.7 atm.

Step-by-step solution

Identify the balanced chemical equation.

The balanced chemical equation is given by: \[ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \]

Calculate the amount of reactant in moles.

We are given that 150 g of dimethylhydrazine reacts with excess dinitrogen tetroxide. First, find the molar mass of dimethylhydrazine to convert grams to moles. The molar mass of dimethylhydrazine: \(M_{(CH_3)_2N_2H_2} = 2(12.01) + 6(1.01) + 2(14.01) + 2(1.01) = 60.12 g/mol\) Now, convert the mass of dimethylhydrazine into moles: \(\mathrm{moles\ of\ dimethylhydrazine} = \frac{150g}{60.12 g/mol} = 2.50 mol\)

Calculate the moles of nitrogen gas produced.

Using stoichiometry, for every mole of dimethylhydrazine, 3 moles of nitrogen are produced. So, we can calculate the moles of nitrogen produced: \(\mathrm{moles\ of\ N_2} = 2.50\ mol\ (CH_3)_2N_2H_2 * \frac{3\ mol\ N_2}{1\ mol\ (CH_3)_2N_2H_2} = 7.50\ mol\ N_2\)

Calculate the partial pressure of nitrogen gas.

Using the Ideal Gas Law: \(PV = nRT\). We can calculate the partial pressure of nitrogen gas, using the moles of nitrogen and the given temperature and volume. \(T = 127^{\circ}C + 273.15 = 400.15 K\) \(V = 250 L\) \(R = 0.0821 \,\mathrm{L \cdot atm / (mol \cdot K)}\) Now, we can solve for the pressure: \(P_{N_2}=\frac{n_{N_2}RT}{V} = \frac{7.50\, mol \cdot 0.0821 \,\mathrm{L \cdot atm / (mol \cdot K)} \cdot 400.15 K}{250 L} = 9.89 \,atm\)

Calculate total moles of gas and total pressure.

First, we need to calculate the moles of complete products produced. Using stoichiometry, we know that for every mole of dimethylhydrazine reacted, we produce 4 moles of water vapor and 2 moles of carbon dioxide. \(\begin{aligned} \mathrm{moles\ of\ H_2O}&=2.50\ mol\ (CH_3)_2N_2H_2 * \frac{4\ mol\ H_2O}{1\ mol\ (CH_3)_2N_2H_2}=10.0\ mol\ H_2O \\ \mathrm{moles\ of\ CO_2}&=2.50\ mol\ (CH_3)_2N_2H_2 * \frac{2\ mol\ CO_2}{1\ mol\ (CH_3)_2N_2H_2}=5.0\ mol\ CO_2 \end{aligned}\) Total moles of gases: \(n_{total} = 7.50\, mol\, N_2 + 10.0\, mol\, H_2O + 5.0\, mol\, CO_2 = 22.5\, mol\) Now, we can calculate the total pressure using the Ideal Gas Law: \(P_{total} = \frac{n_{total}RT}{V} = \frac{22.5\, mol \cdot 0.0821 \,\mathrm{L \cdot atm / (mol \cdot K)} \cdot 400.15 K}{250 L} = 29.7\, atm\) In summary, the partial pressure of nitrogen gas is 9.89 atm, and the total pressure in the tank is 29.7 atm.

Key concepts

Chemical Stoichiometry

Understanding chemical stoichiometry is crucial for anyone involved in the field of chemistry, especially those working with reactions, like the one powering the Lunar Lander's fuel. Chemical stoichiometry refers to the calculation of the relative quantities of reactants and products in chemical reactions. It's the backbone of determining how much reactant you'd need or how much product you could produce in a given chemical process.

For students looking at rocket fuel chemistry, stoichiometry allows you to predict the amount of each substance needed or produced. In our example, we saw how the balanced equation dictated that for every mole of dimethylhydrazine, three moles of nitrogen gas would be produced. The calculation was straightforward—a simple multiplication of the moles of the starting substance by the ratio given by the coefficients in the balanced equation. This core concept is also why reactants are often mixed in precise amounts to avoid the waste of materials and ensure the maximum yield of desired products.

Ideal Gas Law

The ideal gas law is a fundamental equation that connects pressure, volume, temperature, and the amount of gas. It's expressed as PV = nRT, where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is temperature in Kelvins. In the context of rocket fuel chemistry, the ideal gas law enables us to determine the conditions associated with gaseous products, like nitrogen, water vapor, and carbon dioxide.

In this exercise, we applied the ideal gas law to find out the partial pressure of the nitrogen gas produced after the reaction occurred. The exercise also illustrated how the total pressure in a given volume could be found by considering all the gaseous products, showing how essential this concept is for anyone working with gases in chemical reactions. One crucial thing to remember is that the temperature must always be in Kelvin for the ideal gas law to work correctly. Making this conversion is a small but significant step in ensuring accurate calculations.

Molar Mass Calculation

To work with chemical stoichiometry and the ideal gas law effectively, you first need to know how to calculate molar mass. This is the mass per mole of a substance and is a critical component of chemical calculations. For dimethylhydrazine, the step-by-step solution showed how to find its molar mass by adding the atomic masses of its constituent atoms—carbon, hydrogen, and nitrogen.

Molar mass serves as a bridge between mass in grams and quantity in moles and is used extensively to convert from one to the other. As seen in the rocket fuel problem, calculating moles from a given mass required the molar mass, which then set the stage for everything that followed. When it comes to complex mixtures like rocket fuels, it's essential to have an accurate molar mass to ensure the mixture proportions provide the correct chemical and physical properties necessary for the reaction to proceed efficiently and safely.