# A 25xx10^-3L volume of silver nitrate at 0.500*mol*L^-1 concentration is mixed with a 25xx10^-3*L volume of 0.500*mol*L^-1 volume of magnesium chloride...what mass will precipitate?

Oct 28, 2016

Under $2 \cdot g$ $A g C l$.

#### Explanation:

We need (i) a stoichiometric equation:

$A {g}^{+} + C {l}^{-} \rightarrow A g C l \left(s\right) \downarrow$

And (ii) the equivalent quantities of silver ion and chloride ion.

$\text{Moles of silver nitrate:}$ $25 \times {10}^{-} 3 L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.0125 \cdot m o l$

$\text{Moles of chloride ion:}$ $2 \times 25 \times {10}^{-} 3 L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.0250 \cdot m o l$

Why is there double the quantity of chloride ion?

Silver ion is clearly the limiting reagent. $0.0125 \cdot m o l$ $A g C l$ should precipitate out, i.e. $0.0125 \cdot m o l \times 143.32 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g

We know that most halides are soluble, EXCEPT for silver, mercurous, and lead halides. Silver chloride is exceptionally insoluble, and precipitates from solution with alacrity. Note that unless special conditions are employed, the silver chloride will reduce to metallic silver on standing.