Percent Yield

Question

Baking soda (NaHCO3) can be made in large quantities by the following reaction:

 

NaCl+NH3+CO2+H2O=NaHCO3+NH4Cl

 

If 9.17 g of NaCl reacts with excesses of the other reactants and 3.79 g of NaHCO3 is isolated, what is the percent yield of the reaction?

Answer 1

Ok, so due to various factors, the amount of baking soda you actually end up with does not equal the amount that can theoretically be made based on the amount of NaCl used.

 

So, you have to convert the 9.17 g NaCl to moles of NaCl:

 

9.17 g NaCl * (1 mol NaCl / 58.44 g NaCl) = 0.157 mol NaCl

 

The reaction above is already balanced, so you can see that it takes 1 mol NaCl to make 1 mol NaHCO3.  So 0.157 mol NaCl would theoretically give you 0.157 mol NaHCO3.  Now you need to convert this to grams of NaHCO3.

 

0.157 mol NaHCO3 * (84.01 g NaHCO3 / 1 mol NaHCO3) = 13.19 g NaHCO3

 

To calculate percent yield, you take how much you actually made as a percentage of how much you theoretically could have made.  

 

3.79 g NaHCO3 / 13.19 g NaHCO3 * 100 = 28.73% yield