H2O = NaHCO3 + NH3Cl if 9.53 g of NaCl reacts with excesses of the other reactants and 4.13
g of NaHCO3 is isolated, what is the percent yield of the reaction?
Solution
moles of NaCl=0.163moles mass of NaHCO3 to be formed
actually=0.163*(84)=13.7g actual mass of NaHCO3 formed=4.13g percentage yield=4.13/13.7
*100= 30.2%