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-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

Find moles of dimethylhydrazine and nitrogen gas

First, we need to find the moles of dimethylhydrazine and the moles of nitrogen gas produced from the reaction. We know that 150 g of dimethylhydrazine is used and the molar mass of dimethylhydrazine is: \([(2\times12)+(6\times1)+(2\times14)+(2\times1)]=46\ \mathrm{g/mol}\) Now we can find the moles of dimethylhydrazine: \(\mathrm{moles\ of\ dimethylhydrazine}=\frac{150\ \mathrm{g}}{46\ \mathrm{g/mol}}=3.26\ \mathrm{mol}\) Using the balanced chemical equation, we can determine the moles of nitrogen gas produced for every mole of dimethylhydrazine reacted: \(1\ \mathrm{mol\ of\ }(CH_3)_2N_2H_2\rightarrow3\ \mathrm{mol\ of\ }N_2\) So, the moles of nitrogen gas produced are: \(3.26\ \mathrm{mol}\times3=9.78\ \mathrm{mol}\)

02

Calculate the partial pressure of nitrogen gas

Now, we can calculate the partial pressure of the nitrogen gas, using the ideal gas law: \(PV=nRT\) Where: \(P=\mathrm{Partial\ pressure\ of\ nitrogen\ gas}\) \(V=250\ \mathrm{L}\) \(n=9.78\ \mathrm{mol}\) \(R=0.08206\ \mathrm{L\ atm/mol\cdot K}\) \(T=127^\circ\mathrm{C}+273.15=400.15\ \mathrm{K}\) Now, solve for the partial pressure (P) of nitrogen gas: \(P_\mathrm{N_2}=\frac{nRT}{V}=\frac{(9.78\ \mathrm{mol})(0.08206\ \mathrm{L\ atm/mol\cdot K})(400.15\ \mathrm{K})}{250 \mathrm{L}}=13.23\ \mathrm{atm}\)

03

Find moles of other gases and total moles in the tank

From the balanced equation, we can find the moles of water and carbon dioxide as follows: \(1\ \mathrm{mol\ of\ }(CH_3)_2N_2H_2\rightarrow4\ \mathrm{mol\ of\ }H_2O\) \(1\ \mathrm{mol\ of\ }(CH_3)_2N_2H_2\rightarrow2\ \mathrm{mol\ of\ }CO_2\) Calculate the moles of water and carbon dioxide produced: \(3.26\ \mathrm{mol}\times4=13.04\ \mathrm{mol\ of\ }H_2O\) \(3.26\ \mathrm{mol}\times2=6.52\ \mathrm{mol\ of\ }CO_2\) Total moles in the tank: \(9.78+13.04+6.52=29.34\ \mathrm{mol}\)

04

Calculate total_pressure in the tank

Now, we can find the total_pressure by using the ideal gas law (PV=nRT) for total moles in the tank: \(P_\mathrm{total}=\frac{n_\mathrm{total}RT}{V}=\frac{(29.34\ \mathrm{mol})(0.08206\ \mathrm{L\ atm/mol\cdot K})(400.15\ \mathrm{K})}{250 \mathrm{L}}=40.01\ \mathrm{atm}\) Thus, the partial pressure of nitrogen gas produced is 13.23 atm, and the total pressure in the tank is 40.01 atm.

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