posted by Tichele on .
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???