Question

Theoretically how many moles of carbonic acid will be produced by the 3.00 g sample of NaHCO3

Answer 1

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.

Answer 2

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