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

Jun 30, 2017

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

${\text{AgNO"_3(aq) + [stackrel(color(blue)(+3))("Cr")("H"_2"O")_4stackrel(color(blue)(-1))("Cl"_2)]stackrel(color(blue)(-1))("Cl")(aq) -> "AgCl"(s) + ["Cr"("H"_2"O")_4"Cl"_2]"NO}}_{3} \left(a q\right)$

Only the ion in the outer coordination sphere gets displaced here; the complex coordinated to the outer ${\text{Cl}}^{-}$ has an overall charge of $+ 1$.

By the stoichiometry of the reaction, the mols of $\text{AgCl}$ are equal to the mols of ${\text{Cl}}^{-}$ 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"))

$= \text{0.01 mols"/cancel"L" xx 100.00 xx 10^(-3) cancel"L}$

$=$ $\textcolor{b l u e}{\text{0.001 mols AgCl}}$