A. Thermal decomposition of 2.765 g NaHCO\(_3\) yields 1.234 g of solid Na\(_2\)CO\(_3\). Calculate the theoretical yield and percent yield of Na\(_2\)CO\(_3\).
B. Thermal decomposition of 2.968 g of a mixture containing NaHCO\(_3\) lost 0.453 g. Calculate the percentage of NaHCO\(_3\) in this unknown mixture.
Answer (3)
2NaHCO3 -> Na2CO3 + H2O + CO2 2.765g NaHCO3/MM = moles NaHCO3 moles NaHCO3 x (1 mole Na2CO3 / 2 moles NaHCO3) x MM Na2CO3 = theoretical yield of Na2CO3
Percent yield is simply the actual yield/theoretical yield (x100 to put it into percentage).
MM = Molar mass (grams of substance per mol)
For A:
To calculate the number of moles, we use the equation:
Number of moles = Molar mass Given mass .....(1)
Given mass of sodium hydrogen carbonate = 2.765 g
Molar mass of sodium hydrogen carbonate = 84 g/mol
Putting values in equation 1, we get:
Moles of sodium hydrogen carbonate = 84 g / m o l 2.765 g = 0.033 m o l
The chemical equation for the thermal decomposition of sodium hydrogen carbonate follows:
2 N a H C O 3 ( s ) → N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g )
By Stoichiometry of the reaction:
2 moles of sodium hydrogen carbonate produces 1 mole of sodium carbonate
So, 0.033 moles of sodium hydrogen carbonate will produce = 2 1 × 0.033 = 0.0165 m o l of sodium carbonate
Now, calculating the mass of sodium carbonate from equation 1, we get:
Molar mass of sodium carbonate = 106 g/mol
Moles of sodium carbonate = 0.0165 moles
Putting values in equation 1, we get:
0.0165 m o l = 106 g / m o l Mass of sodium carbonate Mass of sodium carbonate = ( 0.0165 m o l × 106 g / m o l ) = 1.75 g
To calculate the percentage yield of sodium carbonate, we use the equation:
% yield = Theoretical yield Experimental yield × 100
Experimental yield of sodium carbonate = 1.234 g
Theoretical yield of sodium carbonate = 1.75 g
Putting values in above equation, we get:
% yield of sodium carbonate = 1.75 g 1.234 g × 100 % yield of sodium carbonate = 70.5%
Hence, the percent yield of sodium carbonate is 70.5 %
For B:
To calculate the percentage composition of sodium hydrogen carbonate in mixture, we use the equation:
% composition of sodium hydrogen carbonate = Mass of mixture Mass of sodium hydrogen carbonate × 100
Mass of mixture = 2.968 g
Mass of sodium hydrogen carbonate = 0.453 g
Putting values in above equation, we get:
% composition of sodium hydrogen carbonate = 2.968 g 0.453 g × 100 = 15.26%
Hence, the percent of sodium hydrogen carbonate in the unknown mixture is 15.26 %
The theoretical yield of Na₂CO₃ from the decomposition of 2.765 g of NaHCO₃ is approximately 1.749 g, resulting in a percent yield of about 70.7%. In a mixture weighing 2.968 g that lost 0.453 g, the percentage of NaHCO₃ is approximately 20.6%.
;
