Question

2.4822g of an unknown mixture of sodium carbonate and sodium bicarbonate is mixed with sodium nitrate. 0.644938 liters of CO2 is produced. What is the mass and percent of sodium carbonate in sample?

Answer 1

We have two reactions:

2NaHCO₃ -------> Na₂CO₃ + H₂O + CO₂

NaNO₃ and Na₂CO₃ does not produce carbon dioxide,when heated

assuming the product obtained is at room temperature of 25 ^oC = 298K

moles of CO₂ (n) = PV/RT = 1atm x 0.644938L/0.08206atm.L/mol.K x 298K = 0.026373647 mol

2 mol NaHCO₃ gives 1 mol CO₂

0.026373647 x2 = 0.05275 mol NaHCO₃

molar mass of NaHCO3 = 84.007g/mol

mass = moles x molar mass = 0.05275 mol x  84.007g/mol = 4.43g

It seems there is some error in the question

might be volume of CO₂

If volume is taken = 0.0644938 L

mass of NaHCO3 will come = 0.4431 g

mass of sodium carbonate = 2.4822g - 0.4431g = 2.0391g

% sodium carbonate = 2.0391g / 2.4822g x 100 = 82.15%