Stoicheometry Problem!! HELP

I tried answering the following Q's, but I keep getting the wrong answer:

Na2CO3 + 2HCl -> 2NaCl + H2O + CO2

a) A chemist dissolves an impure sample of Na2CO3, with a mass of 0.250 g, in water. The chemist determines that 30.4 mL of 0.151 M HCl reacts with the Na2CO3 sample. Calculate the percentage purity of the sample.

b) What volume of CO2 is produced, at 21.5 Celsius and a pressure of 104.0 kPa?

*The answers are: a) 97.6 % b) 54.2 mL

PLZ HELP WITH FULL SOLUTIONS & THANK YOU

Lets get the moles of HCl that reacted.
Better still, the millimoles
30.4 x 0.151
that comes to 4.5904
You could say 4.59 x 10-3 moles if you don't like millimoles. (and the spell checker doesn't)

Now, note the equation for the reaction.
Every mole of sodium carbonate needs TWO moles of HCl
So whatever HCl was used, only half as much sodium carbonate was present.
that's 2.295 millimoles
2.295 x 10-3 moles. (of sodium carbonate)
Have you worked out the molar mass of sodium carbonate? 106?
so multiply the molar mass by the moles to give you the grams.
I used millimoles, so I got milligrams. 243 3 of them
which is 0.2433 grams
Which is 97.3 % of the original mass of 0.250 grams.
If that's the wrong answer that you kept getting you've been doing it right all along.

For part b,
the moles of CO2 will be the same number of moles as the sodium carbonate; 2.295 x 10^-3
Use PV =nRT, or V = nRT/P
V = 2.295 x 10^-3 x 8.31 x 294.5/104 = 0054 litres or 54 mL

Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2

(0.0304 L HCl) x (0.151 mol/L) x (1/2) x (105.9886 g Na2CO3/mol) / (0.250 g) =
0.973 = 97.3%

(0.0304 L HCl) x (0.151 mol/L) x (1/2) x (22.4 L/mol) x ((21.5 + 273) / 273)) x
(101.3 / 104.0) = 0.0540 L = 54.0 mL