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 \(150 \mathrm{~g}\) dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(27^{\circ} \mathrm{C}\) in an evacuated 250-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?

Step by step solution

01

Calculate the moles of dimethylhydrazine

First, we need to convert the given mass of dimethylhydrazine to moles. The molar mass of dimethylhydrazine \((\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{H}_{2})\) is approximately \(2 * 12.01 (C) + 6 * 1.01 (H) + 2 * 14.01 (N) = 60.12 \: \mathrm{g/mol}. \) The moles of dimethylhydrazine can be calculated as follows: Moles of dimethylhydrazine = \(\dfrac{150 \: \mathrm{g}}{60.12 \: \mathrm{g/mol}} = 2.5 \: \mathrm{mol}\)

02

Determine the moles of nitrogen gas formed

Using the balanced chemical equation, we can determine the stoichiometric ratio between moles of dimethylhydrazine and nitrogen gas: 1 mol dimethylhydrazine: 3 mol nitrogen gas From Step 1, we know that there are 2.5 moles of dimethylhydrazine. Therefore, the moles of nitrogen gas formed are: Moles of nitrogen gas = \(2.5 \: \mathrm{mol\: dimethylhydrazine} * \dfrac{3 \: \mathrm{mol\: nitrogen\: gas}}{1 \: \mathrm{mol\: dimethylhydrazine}} = 7.5 \: \mathrm{mol\: nitrogen\: gas}\)

03

Calculate the partial pressure of nitrogen gas

Using the ideal gas law (PV=nRT), we can find the partial pressure of nitrogen gas. We are given the volume (V) as 250 L, the number of moles (n) as 7.5 mol, and the temperature (T) as \(27^{\circ}\mathrm{C}\) which needs to be converted to Kelvin (K): T = \(27+273.15 = 300.15 \: \mathrm{K}\) The gas constant (R) is given as 0.0821 L atm/mol K. The ideal gas law then becomes: Partial pressure of nitrogen gas (P) = \(\dfrac{n * R * T}{V} = \dfrac{7.5 \: \mathrm{mol} * 0.0821 \: \mathrm{L\: atm/mol \: K} * 300.15 \: \mathrm{K}}{250 \: \mathrm{L}} = 7.52 \: \mathrm{atm}\)

04

Determine the moles of other gases produced

Using stoichiometry from the balanced chemical equation, we can determine the moles of water vapor and carbon dioxide produced: 2.5 mol dimethylhydrazine: 4 mol water vapor: 2 mol carbon dioxide Moles of water vapor = \(2.5 \: \mathrm{mol\: dimethylhydrazine} * \dfrac{4 \: \mathrm{mol\: water\: vapor}}{1 \: \mathrm{mol\: dimethylhydrazine}} = 10 \: \mathrm{mol\: water\: vapor}\) Moles of carbon dioxide = \(2.5 \: \mathrm{mol\: dimethylhydrazine} * \dfrac{2 \: \mathrm{mol\: carbon\: dioxide}}{1 \: \mathrm{mol\: dimethylhydrazine}} = 5 \: \mathrm{mol\: carbon\: dioxide}\)

05

Calculate the total moles of gas formed

The total moles of gas formed is the sum of moles of nitrogen gas, water vapor, and carbon dioxide: Total moles of gas = 7.5 mol nitrogen gas + 10 mol water vapor + 5 mol carbon dioxide = 22.5 mol

06

Calculate the total pressure in the tank

Using the ideal gas law (PV=nRT), we can find the total pressure in the tank. We have the total moles (n) as 22.5 mol, the volume (V) as 250 L, and the temperature (T) as 300.15 K: Total pressure (P) = \(\dfrac{n * R * T}{V} = \dfrac{22.5 \: \mathrm{mol} * 0.0821 \: \mathrm{L\: atm/mol \: K} * 300.15 \: \mathrm{K}}{250 \: \mathrm{L}} = 22.57 \: \mathrm{atm}\) Thus, the partial pressure of nitrogen gas produced is 7.52 atm, and the total pressure in the tank is approximately 22.57 atm.

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