# 25*g of sodium chloride are mixed with 35.0*g silver nitrate in aqueous solution. How much silver chloride can be prepared?

Oct 29, 2015

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

Note that I assume you are using silver nitrate, $A g N {O}_{3}$.

#### Explanation:

Moles of $A {g}^{+}$, (35.0*g)/(169.9*g*mol^(-1)) = ??

Moles of $C {l}^{-}$, (25.0*g)/(58.4*g*mol^(-1)) = ??

The $N {a}^{+} \left(a q\right)$ and $N {O}_{3}^{-} \left(a q\right)$ ions are spectators and do not participate in the reaction. Chloride anion is clearly in excess; precipitation $A g C l$ is (almost) quantitative. You are free to do the math.

Oct 29, 2015

The maximum yield of $\text{AgCl}$ is $\text{29.5 g}$.

#### Explanation:

Balanced Equation

"NaCl(aq)"+"AgNO"_3("aq")$\rightarrow$$\text{NaNO"_3("aq")"+AgCl(s)}$

This is a limiting reagent question. The reactant that produces the least amount of silver chloride is the limiting reagent, and it determines the amount of silver chloride that can be produced.

For each reactant, determine how much silver chloride can be produced.

1. Determine moles of reactant by dividing given mass of the reactant by its molar mass.

2. Determine moles silver chloride by multiplying moles of the reactant by the mole ratio between the reactant and silver chloride in the balanced equation, with silver chloride in the numerator.

3. Determine mass silver chloride by multiplying moles of silver chloride by its molar mass.

Sodium chloride

(25.0color(red)cancel(color(black)("g NaCl")))/(58.44color(red)cancel(color(black)("g NaCl"))/color(red)cancel(color(black)("mol NaCl")))xx(1color(red)cancel(color(black)("mol AgCl")))/(1color(red)cancel(color(black)("mol NaCl")))xx(143.32"g AgCl")/(1color(red)cancel(color(black)("mol AgCl")))="61.3 g AgCl"

Silver nitrate

(35.0color(red)cancel(color(black)("g AgNO"_3)))/(169.87color(red)cancel(color(black)("g AgNO"_3))/color(red)cancel(color(black)("mol AgNO"_3)))xx(1color(red)cancel(color(black)("mol AgCl")))/(1color(red)cancel(color(black)("mol AgNO3")))xx(143.32"g AgCl")/(1color(red)cancel(color(black)("mol AgCl")))="29.5 g AgCl"

$\text{AgNO"_3}$ is the limiting reagent. The maximum yield of $\text{AgCl}$ is $\text{29.5 g}$.