A 3.23 gram sample of a sodium bicarbonate mixture was reacted with excess acid. 48.52 mL of the evolved gas was collected of the water at .930 atm. and 23.0 degrees C. What is the percentage of sodium bicarbonate in the original mixture?

Answer 1

I would convert 0.930 atm to mm Hg. 0.930*760 = ??
Then use (P1V1/T1) = (P2V2/T2) to convert 48.52 mL to V2 at STP. Remember to correct P1 for vapor pressure of water. Look up v.p. H2O at 23 C and subtract from P1.
Finally, mL CO2 at STP x (1 mol/22,400 mL) converts to moles CO2, that times molar mass CO2 converts to grams and (g/mass sample)*100 = %CO2
Post your work if you get stuck.

Answer 2

I started to use PV=nRT.
So,
n = (.930 atm)(.04852 L)
------------------ = 1.86 x 10^ -3
(.0821) (296 K) mol

I multiplied that by 84 g/mol (molar mass of sodium bicarbonate) and then by 100%....which equated to 15.62%.
That seems off to me. Do you know what I'm doing wrong?

Answer 3

The first thing you did wrong is that you didn't correct for the vapor pressure of water at 23 C. Therefore, the n you calculated is n for moles CO2 + moles H2O whatever that means (it means the gas is wet and not dry). To correct, look up the vapor pressure of water at 23 C (usually given in tables in mm Hg; therefore, convert that to atm and subtract from the 0.930 atm if you want to continue with your method.) Then recalculate n, which this time will be for the dry gas. The n you have should be a little lower than 0.00186 but not by much. The second thing you did wrong is that you calculated grams NaHCO3 (which you should have done) BUT you didn't calculate that as the percent of the 3.23 g sample.
%NaHCO3 = (mass NaHCO3/mass sample)*100 = ?? I think you should end up with some value about 4 or 5%.

Please note that if you worked it as I did, that I calculated %CO2 and not %NaHCO3. There is nothing wrong with what I did; however, I didn't answer the question (at least not completely).
I hope this helps.