# Sally and Sam reacted 20.0 mL of 1.0 M silver (I) chloride with metallic zinc in class. Sam and Sally collected 0.85 g of zinc chloride. What is the percent yield?

Feb 29, 2016

62%

#### Explanation:

Start by writing a balanced chemical equation for this single replacement reaction

${\text{Zn"_text((s]) + color(red)(2)"AgCl"_text((aq]) -> "ZnCl"_text(2(aq]) + 2"Ag}}_{\textrm{\left(s\right]}}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between silver chloride and zinc chloride. This tells you that the reaction should theoretically produce $1$ mole of zinc chloride for every $\textcolor{red}{2}$ moles of silver chloride that take part in the reaction.

Now, the problem doesn't provide information about the mass of zinc metal used in the experiment, so you can just assume that zinc is In excess.

This means that all the moles of silver chloride will take part in the reaction.

Use the solution's molarity to determine how many moles you have in that sample

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{A g C l} = \text{1.0 M" * 20.0 * 10^(-3)"L" = "0.020 moles AgCl}$

The reaction should theoretically produce

0.020 color(red)(cancel(color(black)("moles AgCl"))) * "1 mole ZnCl"_2/(color(red)(2)color(red)(cancel(color(black)("moles AgCl")))) = "0.010 moles ZnCl"_2

To convert this to grams, use zinc chloride's molar mass

0.010 color(red)(cancel(color(black)("moles ZnCl"_2))) * "136.286 g"/(1color(red)(cancel(color(black)("mole ZnCl"_2)))) = "1.363 g"

So, this tells you that the reaction's theoretical yield, which is what you get in the hypothetical case in which a reaction has a 100% yield, is equal to $\text{1.363 g}$ of zinc chloride.

However, you only recover $\text{0.85 g}$ of zinc chloride. This is the actual yield of the reaction.

The percent yield will thus be

$\textcolor{b l u e}{\text{% yield" = "actual yield"/"theoretical yield} \times 100}$

Plug in your values to get

"% yield" = (0.85color(red)(cancel(color(black)("g"))))/(1.363color(red)(cancel(color(black)("g")))) xx 100 = color(green)("62%")

The answer must be rounded to two sig figs, since that is how many sig figs you have for the molarity of the silver chloride solution and for the mass of zinc chloride recovered.