Gas Stoichiometry & Ideal Gas Law I need a little help with one of the problems below. Any help on the problem below will be appreciated. I don't want to sound like an idiot, I will also provide my work.


  Solid $\ce{NaHCO3}$ decomposes to form water vapor, carbon dioxide and sodium carbonate.
  $$\ce{NaHCO3 -&gt; H2O + CO2 + Na2CO3}$$
  If $2.00\ \mathrm g$ of $\ce{NaHCO3}$ decompose at $500\ \mathrm{^\circ C}$ and $700\ \mathrm{mmHg}$, how many liters of $\ce{CO2}$ should be produced?


$$\mathrm{\frac{750\ \mathrm{Torr}}{760\ \mathrm{Torr}}=0.99\ \mathrm{atm}}$$
$$pV=nRT$$

$$2.00\ \mathrm g\ \ce{NaHCO3} \cdot \frac{1\ \mathrm{mol}\ \ce{NaHCO3}}{84.02\ \mathrm g} \cdot \frac{1\ \mathrm{mol}\ \ce{CO2}}{2\ \mathrm{mol}\ \ce{NaHCO3}} \cdot \frac{22.4\ \mathrm L \ce{CO2}}{1\ \mathrm{mol}\ \ce{CO2}} = 0.287\ \mathrm L\ \ce{CO2}$$

$$0.99\ \mathrm{atm}\cdot0.287\ \mathrm L=n\cdot773\ \mathrm K\cdot0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$$
$$n = 0.004159\ \mathrm{mol}\ \ce{CO2}$$

$$0.004159\ \mathrm{mol}\ \ce{CO2} \cdot\frac{22.4\ \mathrm L}{1\ \mathrm{mol}\ \ce{CO2}} = 0.09317\ \mathrm{L}\ \ce{CO2}$$
I do not like to post an answer for my own question, however I feel the need to clarify my mistakes ton anyone else. So here is the answer:


Calculate the moles of $\ce{NaHCO3}$
Then I need to plug in the moles to $\ce{PV = nRT}$
After finding the volume of $\ce{NaHCO3}$, now I plug it in a dimensional analysis to figure out the volume of $\ce{CO2}$. That is all and the answer would be $\ce{.77\ Liters\ CO2}$