Question

A factory running the Solvay process analyzes its efficiency as follows: Every 4.30 tonnes of reacted NaCl forms 4.75 tonnes of NaHCO3. What is the % yield of this process in the conversion of NaCl to NaHCO3?

Answer 1

Reaction for Solvay Process: NH4NO3 + NaCl NaHCO3 + NH4Cl

1 mol NaCl gives 1 mol of NaHCO3.

now, 1 ton= 907185 gm

4.3 ton NaCl=(4.3*907185)gm = 3900895.5 gm= 66750.4 mol of NaCl [ Molar mass of NaCl=58.44 gm/mol]

4.75 ton NaHCO3=(4.75*907185)gm=4309128.8 gm=51299.2 mol of NaHCO3 [ Molar mass of NaHCO3=84gm/mol]

Theoretically, 66750.4 mol of NaCl should give 66750.4 mol of NaHCO3

Theoritical yeild = 66750.4 mol and Actual yield= 51299.2 mol

Formula for % yeild=

% yield of this process in the conversion of NaCl to NaHCO3= (51299.2/66750.4)*100 = 76.8 %