Question

A 2gm sample containing sodium carbonate and sodium bicarbonate loses .248 gm when heated at 300 degree Celsius, the temperature at which sodium carbonate decomposes to produce sodium carbonate and water. What is the percentage of sodium carbonate in the mixture.

Answer 1

Dear Student,
The balanced equation for sodium bicarbonate is as follow,
2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
Molecular weight of sodium bicarbonate is 84 gm, molecular weight of sodium carbonate is 106 gm, molecular weight of carbon dioxide is 44 gm, and molecular weight of water is 18gm.
Two moles sodium bicarbonate's molecular weight is 168 gm.
For two moles of sodium bicarbonate it loses mole's water and one mole;s carbon dioxide weight. So it lose (44 + 18) gm = 62 gm weight.
At 300 degree celcius,
Here weight loses = 0.248 gm
So reacted sodium bicarbonate in the mixture is 0.248*168/62 = 0.672 gm
168 gm sodium bicarbonate give 106 gm sodium carbonate.
So 0.672 gm sodium bicarbonate give 106*0.672/168 = 0.424 gm sodium carbonate.
So total weight consumed is (0.672 + 0.424) gm = 1.096 gm
Now the remaining weight = (2 - 1.096) gm = 0.904 gm.
So percentage of sodium carbonate in the mixture is 0.424*100/(2 - 0.248) = 42.4/1.752 = 24.2
So 24.2% is the sodium carbonate's presence in the mixture.
Regards