How many grams of silver chloride can be prepared by the reaction of 105.0 ml of 0.22 m silver nitrate with 105.0 ml of 0.13 m calcium chloride?

The balanced equation for the above reaction is as follows;
2AgNO₃ + CaCl₂ ---> 2AgCl + Ca(NO₃)₂
stoichiometry of AgNO₃ to CaCl₂ is 2:1
we first need to find the limiting reactant 
number of moles reacted = molarity x volume
number of AgNO₃ moles = 0.22 mol/L x 0.1050 L = 0.023 mol
number of CaCl₂ moles = 0.13 mol/L x 0.1050 L = 0.014 mol
according to molar ratio of 2:1
if we assume AgNO₃ to be the limiting reactant 
if 2 mol of AgNO₃ react with 1 mol of CaCl₂
then 0.023 mol of AgNO₃ reacts with - 0.023/2 = 0.012 mol of CaCl₂
0.012 mol of CaCl₂ is required but 0.014 mol of CaCl₂ is required
therefore CaCl₂ is in excess and AgNO₃ is therefore the limiting reactant 

the amount of products formed depends on the amount of limiting reactant present 
stoichiometry of AgNO₃ to AgCl is 2:2
the number of moles of AgCl formed = number of AgNO₃ moles reacted 
therefore number of AgCl moles formed = 0.023 mol
mass of AgCl formed = 0.023 mol x 143.3 g/mol = 3.3 g
mass of AgCl formed = 3.3 g

Ibelieve the answer is d