If 13.45 mg of silver chloride were collected at the end of the reaction , how many milliliters of the 0.5558 M magnesium chloride solution reacted ?

Question

Consider the following reaction

MgCl_2(Aq) + 2 AgNO_3(Aq) -> 2AgCl(s) + Mg(NO₃)_2(Aq)

Answer 1

MgCl_2(Aq) + 2 AgNO_3(Aq) ==> 2AgCl(s) + Mg(NO₃)_2(Aq) ,,, balanced equation

molar mass AgCl = 143.3 g/mol

moles AgCl formed in the reaction = 13.45 mg x 1 g / 1000 mg x 1 mol / 143.3 g = 9.386x10^-5 mols

moles MgCl₂ needed to react = 9.386x10^-5 mol AgCl x 1 mol MgCl₂ / 2 mol AgCl = 4.693x10^-5 mol MgCl₂

Volume of 0.5558 M needed = (x L)(0.5558 mol/L) = 4.693x10^-5 mol and x = 8.444x10^-5 L = 0.08444 mls**

**So, if the problem really meant to say that 13.45 mg of silver chloride were collected (and not 13.45 g), then this should be the correct solution to the problem.