Calculate the percentage of NaHCO3 in this unknown mixture.?

Thermal decomposition of 2.968g of a mixture containing NaHCO3 lost 0.453g. Calculate the percentage of NaHCO3 in this unknown mixture.Calculate the percentage of NaHCO3 in this unknown mixture.?
thermal decomp of NaHCO3 occurs as follows



2NaHCO3 --------%26gt;Na2CO2(s) + H2O(g) + CO2(g)



CO2 and H2O are evolved as gases, so the mass lost is the mass of CO2 and H2O produced.



The balanced equation shows that H2O and CO2 are produced in a 1:1 ratio. So the 0.453 g lost represents equal moles of H2O and CO2



Thus let n = moles

n x molar mass H2O + n x molar mass CO2 = 0.453

18.016n + 44.01n = 0.453

62.026 n = 0.453

n = 0.453 / 62.026

= 0.007303 moles of H2O and CO2 are formed



2 moles NaHCO3 decomposes to form 1 mole of H2O

Therefore moles NaHCO3 = 2 x moles H2O

= 0.014607 mol NaHCO3



mass NaHCO3 = molar mass x moles

= 84.008 g/mol x 0.014607 mol

= 1.227 g



% mass = actual mass / total mass x 100/1

= 1.227 g / 2.968 g x 100/1

= 41.34 %

= 41.3 % (3 sig figs)