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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. $$\ce{NaHCO3 -> H2O + CO2 + Na2CO3}$$ 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?

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

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

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

$$0.004159\ \mathrm{mol}\ \ce{CO2} \cdot\frac{22.4\ \mathrm L}{1\ \mathrm{mol}\ \ce{CO2}} = 0.09317\ \mathrm{L}\ \ce{CO2}$$

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  • $\begingroup$ You can't use 22.4 L of $CO_{2}$ since the system is not at STP. It is not needed, however, and you can stop your stoichiometry at the moles of gas produced then use the gas law to solve for V, since everything else is provided. $\endgroup$ – bobthechemist Dec 7 '14 at 14:40
  • $\begingroup$ @bobthechemist So what you are saying is that after figuring out the number of moles of CO2 then you just plug it in the the Ideal Gas Law as 'n' to calculate the number of Liters of CO2. Ok now my other question is if you figure out the liters first of CO2 and then plug it in to the 'V', why I am I getting two different answers? One for solving for moles and one for solving for Liters of CO2 and then plugging it in the equation. Is there some sort of mistake I am making? Thanks for your help. $\endgroup$ – Asker123 Dec 7 '14 at 15:42
  • $\begingroup$ To your first question, yes. For the second, take a closer look at my comment again; you cannot figure out the volume first. Your 22.4 L/mol conversion factor is only true at STP, which is not true in this problem. $\endgroup$ – bobthechemist Dec 7 '14 at 16:10
  • $\begingroup$ For dimensional analysis the main function to use is the fraction. Take a look at my edit. $\endgroup$ – John Snow Dec 7 '14 at 16:37
  • $\begingroup$ I don't get how dividing 750 Torr by 760 Torr you can obtain a number with pressure dimension unit. This .99 should be unit free to be strictly rigorous. $\endgroup$ – Babounet Dec 8 '14 at 8:28
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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:

  1. Calculate the moles of $\ce{NaHCO3}$
  2. Then I need to plug in the moles to $\ce{PV = nRT}$
  3. 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 $\ce{.77\ Liters\ CO2}$
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  • $\begingroup$ In step 2, do you plug in the moles of $\ce{NaHCO3}$? The law is a gas law and $\ce{NaHCO3}$ is a solid. $\endgroup$ – Del Pate May 18 '15 at 7:00

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