2NaHCO3 ---> Na2CO3 + H2O + CO2
Moles of NaHCO3 = 9.3/84 =0.1107 moles
Mole Ratio ---> 2:1
Thus, Moles of CO2 formed(theoretically) = 0.1107/2 =0.055 moles
And, Moles of CO2 formed(practically) = 2.24/44 = 0.0509 moles
Thus % Purity = (0.0509/0.055)*100
= 91.96% = 92%