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:

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:

    .....(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:

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

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 = 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:

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

Experimental yield of sodium carbonate = 1.234 g

Theoretical yield of sodium carbonate = 1.75 g

Putting values in above equation, we get:

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:

Mass of mixture = 2.968 g

Mass of sodium hydrogen carbonate = 0.453 g

Putting values in above equation, we get:

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