chemistry question, pls help me out

When 2.00 mol of the acid react with sodium carbonate, 1.00 mol of carbon dioxide
is formed.
Calculate the volume of carbon dioxide formed, in cm3, from 8.00 g of the acid at
room temperature and pressure.

gonna need the full question for this one I'm pretty sure

gonna need the full question for this one I'm pretty sure

1) Percentage of Oxygen is 100-40-6.67 = 53.33
Therefore if you divide each of the percentages by the relative atomic masses (40/12 , 6.67/1 , 53.33/16) you get 3.33:6.67:3.33 (C : H : O)
Which in its simplest form is 1:2:1, hence the empirical formula is 1:2:1 (CH2O)
You're told the Mr of the acid is 60 g/mol hence since the empirical formula is 30g/mol you can see that the empirical formula needs to be scaled by a factor of 2 so that its the same as the molecular formula. Hence the molecula formula is C2H4O2 (or ethanoic acid)
2) Total number of atoms in 6g of acid is therefore 6/60 moles of acid (0.1) which means the total number of molecules of acid is 0.1 x avogadro number = 6.022 x 10^22, hence the total number of atoms is this x8 since there are 8 atoms in each molcule
3)2 mol of acid produces 1 mol CO2. 2 mol acid is the same as 120g of acid hence if we used 8 grams of acid this is x15 less acid so the number of moles of CO2 produces will be fifteen times less (or 1/15)
And multiplying this by 24 (because there are 24 mol per dm^-3 at RTP) we can say that there must be 1.6dm^3 CO2 which is 1600cm^3

That should help.