Question

4g of a mixture of Na₂CO₃ and NaHCO₃ on heating liberates 448 mL of CO₂ at STP. The percentage of Na₂CO₃ in the mixture is?

Answer 1

Dear Student,

Sodium bicarbonate decomposes in the following manner:
2NaHCO3 = Na2CO3 + H2O + CO2

Sodium carbonate is stable and does not dissociate upon normal heating.

Now,
Volume of CO2 = 448 ml
Volume of 1 mol of gas = 22.4 l
So number of moles of CO2=0.448/22.4 = 0.02 mol

Now,
The molar ratio of NaHCO3 and CO2 is 2:1.
So for 0.02 mol of CO2, 0.04 mol of NaHCO3 are required.
Now, Molar mass of NaHCO3 =84 g mol-1
So, weight of NaHCO3 = number of moles x molar mass = 3.36 g required

Molar ratio for CO2:H2O: Na2CO3 = 1:1:1

Now, weight of CO2 produced = 0.02 x 44=0.88 g
Weight of H2O produced = 0.02 x 18=0.36 g
Weight of Na2CO3 produced = 0.02 x 106 = 2.12 g
Initial amount of Na2CO3= 1-3.36 = 0.64

So, percentage of Na2CO3 in the mixture is (2.12+0.64)/4 =0.69 = 69 % (assuming complete dissociation of sodium bicarbonate takes place)