What is the exp yld when 18.3mL of a .6 M sol of iron(3) chloride is combined with 18 mL of a .679 M sol of silver nitrate at a .825 yield?

Initial number of moles of FeCl₃ = (0.6 mol/L) × (18.3/1000 L) = 0.01098 mol
Initial number of moles of AgNO₃ = (0.679 mol/L) × (18/1000 L) = 0.01222 mol

FeCl₃(aq) + 3AgNO₃(aq) → 3AgCl(s) + Fe(NO₃)₃(aq)
Mole ratio FeCl₃ : AgNO₃ = 1 : 3

If 0.01222 mol AgNO₃ completely reacted, FeCl₃ needed = (0.01222 mol) × (1/3) = 0.00407 mol < 0.01098 mol
Hence, FeCl₃ is in excess, and AgNO₃ completely reacts (limiting reagent/reactant).

According to the above equation, mole ratio AgNO₃ : AgCl = 3 : 3 = 1 : 1
Maximum no. of moles of AgCl formed = 0.01222 mol

Molar mass of AgCl = (107.9 + 35.5) g/mol = 143.4 g/mol
Experimental yield of AgCl = (0.01222 mol) × (143.4 g/mol) × 82.5% = 1.45 g

** n ( FeCl3 ) = 18.3 ml x ( 0.6 mol/ 1000 ml) = 0.011 mol
** n ( AgNO3 ) = 18ml x ( 0.679 mol / 1000 ml) = 0.0122 mol : limiting reactant.
FeCl3 + 3 AgNO3 --------> 3 AgCl ( s) + Fe(NO3 )3
Since AgNO3 isa limiting reactant, 0.0122 mol AgNO3 reacts with FeCl3 to produce 0.0122 mol AgCl (theoretical yield ). But thepercent yield is only 82.5 %, therefore the amount of AgCl obtains : 0.0122 mol x 0.825 = 0.01 mol. The mass of AgCl :
0.01 mol x ( 143.5 g / mol ) = 1.44 g