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
