In: Chemistry
the balanced reaction for both are:
2NaHCO3 ------------> Na2O + 2 CO2 + H2O
Na2CO3 ---------------> Na2O + CO2
From two reactions we can clearly observe that only in reaction-1 water is produced hence 0.0357 moles of water is produced from reaction-1. Observing their co-efficients, when 2 moles of CO2 are produced 1 mole of H2O is produced.
Therefore when 0.0357 moles of H2O is produced, 2*0.0357 = 0.0714 moles of CO2 is produced.
also amount of NaHCO3 present = 2*0.0357 = 0.0714 moles=0.0714*84 =5.99 grams
Remaining CO2 is produced from other equation.
Amount of CO2 released from Na2CO3 = 0.1091 - 0.0714 =0.0377 moles
amount of Na2CO3 = 0.0357 moles = 0.0377*105.9 =3.99 grams
ratio of NaHCO3 to Na2CO3 = 5.99/3.99 = 1.5
pressure=743 torr = 743/760 atm =0.977 atm
temperature = 43 degrees = 43+273 = 316 K
Ideal gas equation PV =nRT
volume of CO2 released by NaHCO3 V = nRT/P = 0.0714*0.0821*316/0.977 = 1.895 lit
volume of CO2 released by Na2CO3 V = nRT/P = 0.0377*0.0821*316/0.977 = 1.001 lit