If "100 mL" of "0.01 mol/L" tetraaquadichlorochromium(III) chloride reacts with excess silver nitrate, how many mols of silver chloride get produced?

1 Answer
Jun 30, 2017

This is just a double replacement reaction in disguise. The balanced reaction has all coefficients of 1:

"AgNO"_3(aq) + [stackrel(color(blue)(+3))("Cr")("H"_2"O")_4stackrel(color(blue)(-1))("Cl"_2)]stackrel(color(blue)(-1))("Cl")(aq) -> "AgCl"(s) + ["Cr"("H"_2"O")_4"Cl"_2]"NO"_3(aq)

Only the ion in the outer coordination sphere gets displaced here; the complex coordinated to the outer "Cl"^(-) has an overall charge of +1.

By the stoichiometry of the reaction, the mols of "AgCl" are equal to the mols of "Cl"^(-) from the limiting reagent, which is implicitly declared to be the transition metal complex, ["Cr"("H"_2"O")_4"Cl"_2]"Cl".

color(blue)(n_(AgCl) = n_(["Cr"("H"_2"O")_4"Cl"_2]"Cl"))

= "0.01 mols"/cancel"L" xx 100.00 xx 10^(-3) cancel"L"

= color(blue)("0.001 mols AgCl")