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

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

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Answer 1 · 2017-06-30T03:39:14

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

${AgNO}_{3} (a q) + [+ 3 Cr {(H_{2} O)}_{4} - 1 {Cl}_{2}] - 1 Cl (a q) \to AgCl (s) + [Cr {(H_{2} O)}_{4} {Cl}_{2}] {NO}_{3} (a q)$ $"AgNO"_3(aq) + [stackrel(color(blue)(+3))("Cr")("H"_2"O")_4stackrel(color(blue)(-1))("Cl"_2)]stackrel(color(blue)(-1))("Cl")(aq) -> "AgCl"(s) + ["Cr"("H"_2"O")_4"Cl"_2]"NO"_3(aq)$

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

By the stoichiometry of the reaction, the mols of $AgCl$ $"AgCl"$ are equal to the mols of ${Cl}^{-}$ $"Cl"^(-)$ from the limiting reagent, which is implicitly declared to be the transition metal complex, $[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl$ $["Cr"("H"_2"O")_4"Cl"_2]"Cl"$ .

$n_{A g C l} = n_{[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl}$ $color(blue)(n_(AgCl) = n_(["Cr"("H"_2"O")_4"Cl"_2]"Cl"))$

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

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

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

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

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