Respuesta :

NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol 
Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% 
3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3 

0.8211 grams Na + 1.266 grams Cl = 2.087 grams
transitionstate

Answer:

                2.21 g of H₂CO₃

Solution:

              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.

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