Posted by Tichele on Saturday, December 11, 2010 at 9:53pm.
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???
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