At elevated temperature, NaHCO3 is concerted quantitatively to Na2CO3. Ignition o a 0.3592 g sample of an antacid tablet containing NaHCO3 ?
And non volatile impurities yielded a residue weighing 0.2362 g. Calculate the % purity of the NaHCO3 sample.
+ CO2(g) + H2O(g)
- Roger the MoleLv 72 months agoFavorite Answer
(0.3592 g total) - (0.2362 g solids) = 0.123 g CO2 & H2O gases
2 NaHCO3(s) → CO2(g) + H2O(g) + Na2CO3(s)
CO2 and H2O are produced in equal numbers of moles, so they can be considered as a single compound, CH2O3.
(0.123 g CH2O3) / (62.0248 g CH2O3/mol) x (2 mol NaHCO3 / 1 mol CH2O3) x
(84.0066 g NaHCO3/mol) / (0.3592 g total) = 0.9276 = 92.76% NaHCO3 by mass