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.

To determine the volume of CO2 evolved at a given pressure and temperature, we need to use the ideal gas law equation: PV = nRT.

First, let's find the number of moles of CO2 produced during the reaction. We know that one mole of NaHCO3 produces one mole of CO2, and one mole of Na2CO3 produces one mole of CO2 as well.

Let's start by calculating the number of moles of NaHCO3 and Na2CO3 in the mixture:
- The molar mass of NaHCO3 is calculated by adding the atomic masses of sodium (Na), hydrogen (H), carbon (C), and three oxygen atoms (O):
23 + 1 + 12 + (3 * 16) = 84 g/mol.
- The molar mass of Na2CO3 is calculated by adding the atomic masses of sodium (Na), carbon (C), and three oxygen atoms (O):
(2 * 23) + 12 + (3 * 16) = 106 g/mol.

To find the moles of NaHCO3 and Na2CO3, we divide the mass of each compound by their respective molar masses:
- Moles of NaHCO3 = 1.600 g / 84 g/mol = 0.0190 mol.
- Moles of Na2CO3 = 1.600 g / 106 g/mol = 0.0151 mol.

Since the reaction completely converts both compounds to CO2, the total moles of CO2 produced will be the sum of the moles of NaHCO3 and Na2CO3:
Total moles of CO2 = 0.0190 mol + 0.0151 mol = 0.0341 mol.

Now, we can use the ideal gas law equation to find the volume of CO2 at the given pressure and temperature.
- Pressure (P) = 0.984 mm Hg = 0.984 torr. We need to convert this to atm by dividing by 760 (since 760 torr = 1 atm):
P = 0.984 torr / 760 torr/atm = 0.001293 atm.
- Temperature (T) = 25 degrees Celsius = 298 degrees Kelvin.

Using the ideal gas law equation PV = nRT:
V = (nRT) / P, where V is the volume of CO2.

Substituting the known values:
V = (0.0341 mol * 0.0821 L·atm/mol·K * 298 K) / 0.001293 atm.
V ≈ 6410 mL, or 6.41 L.

Therefore, the volume of CO2 evolved at 0.984 mm Hg and 25 degrees Celsius is approximately 6410 mL, or 6.41 L.