A mixture of $\ce{NH3}$ and $\ce{N2H4}$ is placed in a sealed container at $\pu{300 K}$. The pressure within the the container is $\pu{0.6 atm}$. The container is heated to $\pu{1000 K}$ where the gases undergo decomposition reactions. The pressure at this stage becomes $\pu{4.8 atm}$. What was the mole percent of $\ce{NH3}$ in the original mixture?

Decomposition reactions:

$$ \begin{align} \ce{2 NH3 &–> N2 + 3 H2}\\ \ce{N2H4 &-> N2 + 2 H2} \end{align} $$

Let the partial pressure of $\ce{NH3}$ be $p_a$ and the partial pressure of $\ce{N2H4}$ be $p_b$. Then

$$p_a + p_b = 0.6$$

Using Gay-Lussac's law,

$$p\propto T$$

Can we work on this problem? I'm not able to proceed further.

  • $\begingroup$ You first calculate the pressure after heating 300k gas at 0.6 atm to 1000k using pv=nrt. You will need to multiply the result by a factor to get to the final pressure. This factor is the final number of moles divided by the starting number of moles (notice the decomposition reactions increase the nuber of moles). Fiddle around with the proportions of both gases until you get the right amount of moles before and after heating. $\endgroup$ – Francis L. Jul 31 '19 at 23:21

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