A 3.23 gram sample of a sodium bicarbonate mixture was reacted with excess acid. 48.52 mL of the evolved gas was collected over water at 0.930 arm and 23.0 degrees Celsius. What is the percentage of sodium bicarbonate in the original mixture?
The pressure of water at 23.0 degrees Celsius is 0.028 atm
Again, any help would be greatly appreciated!
Use PV = nRT and solve for n using the conditions in the problem. P that you substitute (in atm) is 0.930-(23.0/760) to correct for the fact it is collected over water and isn't dry. Then convert moles CO2 to moles NaHCO3 to grams NaHCO3 to percent NaHCO3.
This might be a stupid question, but why do I convert to moles of CO2 first?
Ok, so I've worked out the problem but I'm getting something like 6.6 x 10^-4 percent. That can't be right, can it?
Here's my work:
NaHCO3(s) + HCl(aq) ---> NaCl(aq) + CO2(g) + H2O(l)
Pressure of the gas= pressure of atmosphere - pressure of 23.0 degrees Celsius (found from table in my chem book) this comes out to .930atm-0.0277atm= 0.902atm
moles of CO2:
n=PV/RT= (.902atm x (48.53 x 10^-3 mL))/ (.08206 x 296)= 0.00180 moles CO2
Moles of NaHCO3:
0.00180mol CO2 x 1 mol NaHCO3/ 1 mol CO2= 0.00180 moles NaHCO3
Mass of NaHCO3:
0.00180mol/84 grams= 2.143 x 10^-5 g NaHCO3
Percent= (2.143x 10^-5) / 3.32 grams all multiplied by 100
Here is where I'm getting that 6.6 something x 10^-4 or something percent. Is that even possible???
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 initial mass of the mixture.
First, let's calculate the moles of the evolved gas (CO2) using the ideal gas law:
PV = nRT
Where:
P = partial pressure of CO2 (0.930 atm)
V = volume of CO2 (48.52 mL, which can be converted to L by dividing by 1000)
n = moles of CO2
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature in Kelvin (23.0 degrees Celsius + 273.15)
Using the equation PV = nRT, we can rearrange it to solve for n:
n = PV / RT
Substituting the values:
n = (0.930 atm) * (48.52 mL / 1000 L/mL) / (0.0821 L·atm/K·mol) * (23.0 + 273.15 K)
Next, we need to calculate the moles of NaHCO3.
The balanced chemical equation for the reaction of NaHCO3 with acid is:
2 NaHCO3 + 2 HCl → 2 CO2 + 2 H2O + 2 NaCl
From the equation, we can see that 2 moles of NaHCO3 are required to produce 2 moles of CO2. Therefore, the moles of NaHCO3 in the reaction are equal to half the moles of CO2 evolved.
n(NaHCO3) = n(CO2) / 2
Finally, we can calculate the percentage of NaHCO3 in the original mixture by dividing the moles of NaHCO3 by the initial mass of the mixture and multiplying by 100:
% NaHCO3 = (n(NaHCO3) * M(NaHCO3)) / (mass of mixture) * 100
Where:
M(NaHCO3) = molar mass of NaHCO3 (84.0 g/mol)
mass of mixture = mass of NaHCO3 + mass of impurities
By plugging in the values and following the steps outlined above, you should be able to determine the percentage of sodium bicarbonate in the original mixture.