# Question f2ae2

Jul 25, 2015

The limiting reagent is copper (II) chloride.

#### Explanation:

Start by writing the balanced chemical equation for this double replacement reaction

$A g N {O}_{3 \left(a q\right)} + C u C {l}_{2 \left(a q\right)} \to A g C {l}_{\left(s\right)} \downarrow + C u {\left(N {O}_{3}\right)}_{2 \left(a q\right)}$

The $1 : 1$ mole ratio that exists between silver nitrate and copper (II) chloride tells you that the reaction needs equal numbers of moles of each reactant.

This means that the reactant that has fewer moles present will act as a limiting reagent.

To determine how many moles of each you have, use the two compounds' molar mass

52.2cancel("g") * ("1 mole "AgNO_3)/(169.87cancel("g")) = "0.3073 moles" $A g N {O}_{3}$

and

24.6cancel("g") * ("1 mole "CuCl_2)/(134.45cancel("g")) = "0.1830 moles" $C u C {l}_{2}$

Looks like the copper (II) chloride will be the limiting reagent, which of course means that the silver nitrate will be in excess.

So, the formula of the limiting reagent is $C u C {l}_{2}$.

The maximum amount of silver chloride that the reaction can produce depends on the number of moles of each reactant that actually react.

SInce the limiting reagent will determine how many moles of silver chloride will be produced, you will have

0.1830cancel("moles"CuCl_2) * ("1 mole "AgCl)/(1cancel("mole"CuCl_2)) = "0.1830 moles"# $A g C l$

The mass of silver chloride that contains this many moles is

$0.1830 \cancel{\text{moles") * "143.32 g"/(1cancel("mole")) = color(green)("26.2 g } A g C l}$

After the reaction is complete, you will be left with

${n}_{A g N {O}_{3}} = 0.3073 - 0.1830 = \text{0.1243 moles}$ $A g N {O}_{3}$

The mass of silver nitrate that remains after the reaction will be

$0.1243 \cancel{\text{moles") * "169.87 g"/(1cancel("mole")) = color(green)("21.1 g} A g N {O}_{3}}$