Question

A.) Thermal Decomposition of 2.765 g NaHCO3 yields 1.234g of a solid Na2CO3 . Calculate the theoreticial yield and percent yield of Na2CO3.
B) Thermal decomposition of 2.968 g of a mixture containing NaHCO3 lost 0.453 g . Calculate the percentage of NaHCO3 in this unknown mixture

Answer 1

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:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(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:

$\text{Moles of sodium hydrogen carbonate}=\frac{2.765g}{84g/mol}=0.033mol$

The chemical equation for the thermal decomposition of sodium hydrogen carbonate follows:

$2NaHCO_3(s)\rightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$

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 = $\frac{1}{2}\times 0.033=0.0165mol$ 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:

$0.0165mol=\frac{\text{Mass of sodium carbonate}}{106g/mol}\\\\\text{Mass of sodium carbonate}=(0.0165mol\times 106g/mol)=1.75g$

To calculate the percentage yield of sodium carbonate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of sodium carbonate = 1.234 g

Theoretical yield of sodium carbonate = 1.75 g

Putting values in above equation, we get:

$\%\text{ yield of sodium carbonate}=\frac{1.234g}{1.75g}\times 100\\\\\% \text{yield of sodium carbonate}=70.5\%$

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:

$\%\text{ composition of sodium hydrogen carbonate}=\frac{\text{Mass of sodium hydrogen carbonate}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 2.968 g

Mass of sodium hydrogen carbonate = 0.453 g

Putting values in above equation, we get:

$\%\text{ composition of sodium hydrogen carbonate}=\frac{0.453g}{2.968g}\times 100=15.26\%$

Hence, the percent of sodium hydrogen carbonate in the unknown mixture is 15.26 %

Answer 2

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)