Why is my solution for the partial pressures of two different gases incorrect? I have the chemical equation: $\ce{NH4NO3(s) -&gt; 2H2O (g) + N2O (g)}$. 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.
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.