Answer to Question #228913 in General Chemistry for Usman Ama

Question

Baking soda (NaHCO3) can be made in large quantities by the following reactions:
NaCL(aq) + NH3(aq) + CO2(g) + H2O(l)= NaHCO3(s) + NH4CL(aq)
If 10.0 g of NaCL reacts with excess of the other reactants and 4.2 g of NaHCO3 is isolated, what is the percentage yield of the reaction?

Answer 1

So, you have to convert the 10.0 g NaCl to moles of NaCl:

10 g NaCl * (1 mol NaCl / 58.44 g NaCl) = 0.17112 mol NaCl

4.2 g NaHCO3 / 10 g NaHCO3 * 100 = 42% yield