Gravimetric analysis of a carbonate

[g07l0989]

question
during a gravimetric anaalysis of a carbonate sample ,0.100g sample was titrated with excess acid the resultant carbon dioxide was collected in 40.0cm3 of 0.05 mol/dm3
Ba(OH)3 solution:
MCO3 + 2H=M+CO2
CO2+ Ba(OH)2=BaCO3 +H2O

Ba(OH)2 uses 22 cm3 of a 0.091 mol/dm3 HCl calculate the % of the carbonate is sample??"

well i found the number of moles of the HCl which i compared to the number of moles of BA(OH)2 by ratio the i got the number of moles of the Ba(OH)2 that was initially ther ,i then substrated that answer to from that that was titrated therefore getting the number of moles of caorbin dioxide but i don't know whether i relate the number of moles of CO2 to the M(see equation)or to MCO3.and wat i do after that to get the mass of the carbonate in the sample...please help if you can!thanks

 

[chemisttree]

You have analyzed for CO2 which exists in a 1:1 ratio to carbonate (CO_3^{-2}). Will you report the amount of carbonate as CO_3^{-2} or as M^{+2}CO_3^{-2}. The former is straightforward but you haven't given enough information here to answer the latter.