posted by Monica .
You have 1.600g mixture of NaHCO3 and Na2CO3. You find that 11.60 ml of 2.00M HCl is needed to convert the sample completely to NaCl, H2O and CO2.
NaHCO3 (aq) + HCl (aq) -> NaCl (aq) + H2O (l) + CO2 (g)
Na2CO3 (aq) +2HCl (aq) -> 2NaCl (aq) + H2) (l) + CO2 (g)
What volume of CO2 is evolved at 0.984 mm Hg and 25 degrees Celsius?
I think you must determine first how much NaHCO3 and Na2CO3 constitute the 1.600 g sample. You can do that the following way.
We know it takes 11.6 mL of 2.00 M HCl.
We know grams NaHCO3/84 = moles NaHCO3 and that will take L = moles NaHCO3/2 M to titrate it. To write that in one step and we call g NaHCO3 = x, then
x/(84*2M) = liters to titrate the NaHCO3. A similar expression for Na2CO3 (grams Na2CO3 = 1.600-x) can be written for grams Na2CO3.
moles Na2CO3 = (1.600-x)/106.
It will take twice as much to titrate Na2CO3 (from the equation in the problem) so 2(1.600-x)/(106*2M) which simplifies to (1.600-x)/106. All of that can be combined into one equation.
Liters to titrate NaHCO3 + L to titrate Na2CO3 = liters to titrate from the problem. Therefore,
x/(84*2) + (1.600-x)/106 = 0.0116 L.
Solve for x = g NaHCO3 and 1.600-x = grams Na2CO3. If I didn't goof that is 1.00 g NaHCO3 and 0.600 g Na2CO3 but you need to confirm that. NOW, you can take the grams of each, calculate how much CO2 is released by the reaction, and use PV = nRT to calculate the volume. Take a look at the problem; I have trouble believing you meant 0.984 mm Hg.