3
$\begingroup$

I have the chemical equation: $\ce{NH4NO3(s) -> 2H2O (g) + N2O (g)}$. There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00\ ^\circ\pu{C}$ until the total pressure is $3.70$ atm.

This is what I tried:

Because there are 2 more $\ce{H2O}$ than $\ce{N2O}$ I said that there are $\pu{9.06g}$ of $\ce{H2O}$ and $\pu{3.02g}$ of $\ce{N2O}$.

$$\pu{9.06g}\times \frac{\pu{mol}}{\pu{18.016 g}} = \pu{0.503 mol}\ \ce{H2O}$$ $$\pu{3.02g}\times \frac{\pu{mol}}{\pu{44.02g}}= \pu{0.0686 mol}\ \ce{N_2O}$$

I then found the mole ratio of both compounds: $$X_{H_{2}O}={0.503\over 0.572}=0.879 $$ $$X_{N_{2}O}={0.0686\over 0.572}=0.1199$$

Then I used the relationship between mole ratios and partial pressure $X_i={P_i \over P_{tot}}$, where $P_{tot}=\pu{3.70atm} $:

$$P_\ce{H2O}=(0.879)(\pu{3.70atm})=\pu{3.25atm}$$ $$P_\ce{N2O}=(0.1199)(\pu{3.70atm})=\pu{0.444atm}$$

According to the answer solution for this problem $P_\ce{H2O}=\pu{2.47atm}$ and $P_\ce{N2O}=\pu{1.23atm}$. What am I doing wrong? I can't figure out how to get the correct solution.

$\endgroup$
4
$\begingroup$

You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $\ce{NH4NO3}$, then you have:

$$\ce{12.8 g NH4NO3}\times\frac{\ce{1 mol NH4NO3}}{\ce{80.04 g NH4NO3}}=\ce{0.160 mol NH4NO3}$$

Since $\ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $\ce{H2O}$ and $\ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer

| improve this answer | |
$\endgroup$
7
$\begingroup$

The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).

$$P_{\ce{N2O}} = \frac{1}{3} \times 3.70\ \mathrm{atm} = 1.23\ \mathrm{atm}$$ $$P_{\ce{H2O}} = \frac{2}{3} \times 3.70\ \mathrm{atm} = 2.47\ \mathrm{atm}$$

The bottom line is that chemical equations (i.e. stoichiometry) describe the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.

| improve this answer | |
$\endgroup$
0
$\begingroup$

Some made-up questions are not well thought thru. This looks like an example. Or, maybe it's a trick question. The exact answer desired is not stated: what exactly is to be found? The ratio of partial pressures (if all vapor)? That is absurdly simple and can be stated by looking at the reaction equation. But that can't be the question. The volume of the container? Possibly.

At 25 C, the vapor pressure of H2O will be about 25mm/760mm = 0.033 atm. The rest of the final pressure (3.70 atm) is N2O. Well, approximately. At 1 atm, 25 C, N2O is soluble in water about 1 liter per liter, so at 3.7 atm, multiply the volume of water by 3.7 to get the volume of N2O removed from the gas phase. The volume of water is 2 x 0.16 x 18/1000 = 0.00576 L. The ratio of partial pressures will then be (3.7 - 0.0213)/0.033 = 111.5 (approximately).

The volume of the container? 1 mole of gas at 25 C occupies 22.4 L. Therefore 0.16 moles of NH4NO3 gives 0.16 moles of N2O, which occupy (approximately) 0.16 x 22.4 L = 3.584 L.

You could continue by calculating the volume of liquid water and how much N2O is dissolved, subtract that from the volume of N2O in the gas phase and recalculate the volume. Reiterate until satisfied, or just accept the approximation.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.