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

About 8 gallons of fuel will be used. (8.08)