This is a limiting reactant problem.
We know that we will need a balanced equation, masses, and moles, etc., so let's gather all the information in one place.
#M_text(r):color(white)(mmmmmmmmmmmmmmm)143.32#
#color(white)(mmmmmmm)"NaCl" + color(white)(ll)"AgNO"_3 → "AgCl" + "NaNO"_3#
#"Moles:"color(white)(mmml)"0.0012"color(white)(mll)"0.010 49"#
#"Divide by:"color(white)(mmm)1color(white)(mmmmll)1#
#"Moles rxn:"color(white)(ml)0.0012color(white)(mll)"0.010 49"#
#"Moles of NaCl" = 0.0165 color(red)(cancel(color(black)("L NaCl"))) × "0.7 mol NaCl"/(1 color(red)(cancel(color(black)("L NaCl")))) = "0.0012 mol NaCl"#
#"Moles of AgNO"_3 = 0.0157 color(red)(cancel(color(black)("L AgNO"_3))) × ("0.668 mol AgNO"_3)/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.010 49 mol AgNO"_3#
Identify the limiting reactant
An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.
You simply divide the number of moles by the corresponding coefficient in the balanced equation.
I did that for you in the chart above.
#"AgNO"_3"# is the limiting reactant because it gives the fewest "moles of reaction".
Calculate the theoretical yield
#"Theor. yield" = "0.010 49" color(red)(cancel(color(black)("mol AgCl"))) × (1 color(red)(cancel(color(black)("mol AgCl"))))/(1 color(red)(cancel(color(black)("mol AgNO"_3)))) × ("143.32 g AgCl")/(1 color(red)(cancel(color(black)("mol AgCl")))) = "1.5031 g AgCl"#
Calculate the experimental yield
#"Experimental yield" = 0.773 × "1.5031 g" = "1.16 g"#
