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.
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?
1 answer
1 answer