Consider the reaction 3 ccl4 + 2 sbf3 → 3 ccl2f2 + 2 sbcl3. if 7.30 g of ccl4 react with 5.50 g of sbf3 and 5.10 g of ccl2f2 is actually formed, what is the percent yield?
the balanced equation for the reaction is; 3CCl₄ + 2SbF₃ ---> 3CCl₂F₂ + 2SbCl₃ stoichiometry of CCl₄ to SbF₃ is 3:2 number of CCl₄ moles - 7.30 g/ 154 g/mol = 0.0474 mol number of SbF₃ moles - 5.50 g/ 179 g/mol = 0.0307 mol if CCl₄ is the limiting reactant, 3 mol of CCl₄ reacts with 2 mol of SbF₃ therefore 0.0474 mol reacts with - 2/3 x 0.0474 = 0.0316 mol but only 0.0307 mol of SbF₃ is present therefore SbF₃ is the limiting reactant and CCl₄ is in excess stoichiometry of SbF₃ to CCl₂F₂ is 2:3 if 2 mol of SbF₃ forms 3 mol of CCl₂F₂ then 0.0307 mol of SbF₃ forms - 3/2 x 0.0307 = 0.0461 mol then mass produced - 0.0461 mol x 121 g/mol = 5.58 g but actual yield was 5.10 g percent yield = actual yield / theoretical yield x 100 % = 91.4 % is the percent yield of CCl₂F₂
