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?
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.
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?
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.
To determine the percentage of sodium bicarbonate in the original mixture, we need to calculate the amount of sodium bicarbonate reacted and compare it to the total mass of the sample.
First, we need to find the number of moles of gas evolved using the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given:
Pressure (P) = 0.930 atm
Volume (V) = 48.52 mL = 0.04852 L (convert mL to L)
Temperature (T) = 23.0 degrees C = 23.0 + 273.15 = 296.15 K (convert Celsius to Kelvin)
Ideal Gas Constant (R) = 0.0821 L*atm/(mol*K)
Now, rearranging the Ideal Gas Law equation to solve for the number of moles (n):
n = PV / (RT)
Substituting the given values:
n = (0.930 atm * 0.04852 L) / (0.0821 L*atm/(mol*K) * 296.15 K)
Next, we need to calculate the amount of sodium bicarbonate (NaHCO3) reacted using stoichiometry.
The balanced chemical equation for the reaction between sodium bicarbonate and acid is:
2 NaHCO3 + 2 HCl -> 2 CO2 + 2 H2O + 2 NaCl
This equation tells us that for every 2 moles of NaHCO3 reacted, 2 moles of CO2 are produced.
Since the number of moles of CO2 produced is equal to the number of moles of gas evolved, we can say:
1 mole of CO2 evolved = 1 mole of NaHCO3 reacted
Therefore, the number of moles of NaHCO3 reacted can be calculated as:
moles of NaHCO3 reacted = 0.5 * n
Now we can determine the mass of NaHCO3 reacted using its molar mass. The molar mass of sodium bicarbonate (NaHCO3) is:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
Finally, we can calculate the mass percentage of NaHCO3 in the original mixture:
mass percentage of NaHCO3 = (mass of NaHCO3 reacted / mass of the sample) * 100
The mass of the sample is given as 3.23 grams.
By following the steps outlined above, you can now calculate the percentage of sodium bicarbonate in the original mixture.