Answer :
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)
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)
Answer:
For A: The percent yield of sodium carbonate is 70.5 %
For B: The percent of sodium hydrogen carbonate in the unknown mixture is 15.26 %
Explanation:
- For A:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(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:
[tex]\text{Moles of sodium hydrogen carbonate}=\frac{2.765g}{84g/mol}=0.033mol[/tex]
The chemical equation for the thermal decomposition of sodium hydrogen carbonate follows:
[tex]2NaHCO_3(s)\rightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)[/tex]
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 = [tex]\frac{1}{2}\times 0.033=0.0165mol[/tex] 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:
[tex]0.0165mol=\frac{\text{Mass of sodium carbonate}}{106g/mol}\\\\\text{Mass of sodium carbonate}=(0.0165mol\times 106g/mol)=1.75g[/tex]
To calculate the percentage yield of sodium carbonate, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of sodium carbonate = 1.234 g
Theoretical yield of sodium carbonate = 1.75 g
Putting values in above equation, we get:
[tex]\%\text{ yield of sodium carbonate}=\frac{1.234g}{1.75g}\times 100\\\\\% \text{yield of sodium carbonate}=70.5\%[/tex]
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:
[tex]\%\text{ composition of sodium hydrogen carbonate}=\frac{\text{Mass of sodium hydrogen carbonate}}{\text{Mass of mixture}}\times 100[/tex]
Mass of mixture = 2.968 g
Mass of sodium hydrogen carbonate = 0.453 g
Putting values in above equation, we get:
[tex]\%\text{ composition of sodium hydrogen carbonate}=\frac{0.453g}{2.968g}\times 100=15.26\%[/tex]
Hence, the percent of sodium hydrogen carbonate in the unknown mixture is 15.26 %