Question

NaCl + NH3+CO2+H2O--> NaHCO3 + NH4Cl

if 9.93g of NaCl reacts with excesses of the other reactants and 4.31g of NaHCO3 is isolated, what is the percent yield of the reaction?

Answer 1

Given :

Reaction :

NaCl + NH3+CO2+H2O-- > NaHCO3 + NH4Cl

Mass of NaCl = 9.93 g

Actual mass of NaHCO3 isolated = 4.31 g

Calculation of moles of NaCl

Moles of NaCl = Mass of NaCl / molar mass of NaCl

= 9.93 g / 58.443 g per mol

= 0.1699 mol

Calculation of moles of NaHCO3 by using moles of NaCl

Mol ratio of NaHCO3 : NaCl is 1 : 1

Number of moles of NaHCO3 = Number of moles of NaCl x 1 mol NaHCO3 / 1 mol NaCl

= 0.1699 mol NaHCO3

Calculation of mass (Theoretical yield) of NaHCO3

= Moles of NaHCO3 x molar mass of NaHCO3

= 0.1699 mol NaHCO3 x 84.0059 g per mol

= 14.27 g

Calculation of percent yield :

Percent yield = (Actual mass / theoretical mass ) x 100

= (4.31 g / 14.27 g ) x 100

= 0.3020 x 100

= 30.2 %

Percent yield = 30.2 %