Question

Student Ann Brilliant heated 5.225 g of sodum carbonate/sodium bicarbonate mixture to a constant mass of 4.272 g, Determine the % weight of sodium bicarbonate in the original mixture. Note: employ 2 digits after the decimal place in your answer and do not use units (i.e. %).

Answer 1

Ans:

1. mass of the mixture before heating: Na2CO3+NaHCO3 = 5.225g

2. mass left over after heating/cooling the mixture is: only solid Na2CO3 = 4.272g

3.Loss in mass is due to loss of CO2 and H2O lost.

4. Loss in mass is the difference of mass of mixture before heating and after heating.

ie mass of CO2 and H2O = 5.225g - 4.272g = 0.953g

  Please Note:

This is the reaction that happens on heating the mixture;

2NaHCO3(s)->Na2CO3(s)+H2O(g)+CO2(g)

(After heating and cooling the mixture all of the NaHCO3 in the mixture has converted to solid Na2CO3 and other products formed are gases ie CO2 gas and H2O vapor which will escape out.)

5. According to the reaction:

2NaHCO3(s)->Na2CO3(s)+H2O(g)+CO2(g)

2 x 84g/mol =168g -> ----- +18g/mol + 44g/mol

6. Hence in general: 168g of NaHCO3 loses = 62g [18+44]

x mass of NaHCO3 in the given mixture/decomposed = 0.953g [mass lost in the given mixture]

7. Taking the ratios: 168/x = 62/0.953

168/x = 65.0577

Solving for x,ie

mass of NaHCO3 in the given mixture = x = 168/65.0577 = 2.582g

8. mass of given mixture = 5.225g

mass of NaHCO3 in the given mixture = 2.582g

% NaHCO3 in mixture = 2.582/5.225*100 = 49.42 %

Ans: % NaHCO3 in mixture = 49.42