Theoretically how many moles of carbonic acid will be produced by the 3.00 g sample of nahco3

                2.21 g of H₂CO₃

              The balance chemical equation for the conversion of sodium bicarbonate into carbonic acid is as follow,

                          NaHCO₃  +  HCl     →     H₂CO₃  +  NaCl

According to equation,

        84 g (1 mole) NaHCO₃ produces  =  62 g (1 mol) of H₂CO₃

So,

                3.0 g of NaHCO₃ will produce  =  X g of H₂CO₃

Solving for X,

                      X  =  (3.0 g × 62 g) ÷ 84 g

                       X  =  2.21 g of H₂CO₃

Hence, if the yield is ideal means, 100 % conversion takes place and all product is recovered then the amount of carbonic acid produced will be 2.21 g.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2