When 6.25 g of AgNO3 reacts with 4.12 g of NaCl to form NaNO3 and AgCl, what is the limiting reactant?

Nov 2, 2015

Molecular mass of $A g N {O}_{3}$, $169.83$ $g \cdot m o {l}^{-} 1$; $N a C l$, $58.44$ $g \cdot m o {l}^{-} 1$. Thus there are about twice the quantity of $N a C l$ species.

Explanation:

The reaction (which perhaps I should have written first!) is:

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

The nitrate and sodium ions are simply along for the ride and remain in solution. Note that the silver chloride precipitate would be exceptionally curdy and hard to handle. $A g C l$ would also tend to decompose and darken to $A {g}_{2} O$ and silver metal.

So what are the molar quantities of silver nitrate and sodium chloride in this rxn? For precious metals, why is it a good idea that the precipitating reagent be present in excess?