Quantitative chemistry question... help?

25.00dm3 of 0.100 mol dm-3 sodium hydrogencarbonate solution were titrated with dilute sulfuric acid.

2NaHCO3(aq) + H2SO4(aq) ----> Na2SO4(aq) + 2H2O(l) + 2 CO2(g)

From the equation: 1mol NaHCO3 will produce 1 mol CO2
I have assumed that you made an error in your submission - you have titrated 25.0cm³ and not 25.0dm³ I will use 25.0cm³

Mol NaHCO3 in 25 .0cm³ of 0.10M solution = 25/1000*0.1 = 2.5*10^-3 mol NaHCO3
This will produce 2.5*10^-3 mol CO2

You want to use the gas equation to calculate volume:
PV = nRT
P = 101.325kPa
V = ??
n= (2.5*10^-3)
R = 8.314
T = 273K

101.325 *V = (2.5*10^-3) * 8.314*273
V = 5.674/ 101.325
V = 0.056L

The volume of CO2 produced = 0.056L OR 56cm³.

Note my use of units - If R = 8.314 , then : P must be in kPa - which you have 101.325 - T must be in K , which is 273 and V will be in L , which is what I got.

In passing I can advise that from my experience here in Yahoo and my own students: Students everywhere love to do conversions - most times this is unnecessary and wrong and only leads to errors: Why have you found it necessary to be converting P = 100kPa to 100,000Pa. The equation calls for the units I have used when R = 8.314 as specified by you . In practice , at STP the pressure is 101.325kPa , but possibly your teacher has rounded to 100kPa. I use and teach 101.325kPa = 1.0 atmosphere

Of course there is always the easier option : At STP , 1.0 mol gas has volume 22.4L.
You have 2.5*10^-3 mol
Volume = (2.5*10^-3)*22.4 = 0.056L OR 56mL
Perhaps your teacher wants you to use the gas equation - but I always check my answer using the 22.4L = 1 mol gas at STP - quick and easy.