# Question c429b

Sep 22, 2016

"% yield" = 99.4%

#### Explanation:

Start by writing the balanced chemical equation that describes this synthesis reaction

${\text{Zn"_ ((s)) + "I"_ (2(aq)) -> "ZnI}}_{2 \left(a q\right)}$

You know that $1$ mole of zinc metals reacts with $1$ mole of aqueous iodine solution to produce $1$ mole of zinc iodide.

Now, use the molar masses of the chemical species involved in the reaction to convert the given masses to moles

0.537 color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38 color(red)(cancel(color(black)("g")))) = "0.008214 moles Zn"

2.046 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81color(red)(cancel(color(black)("g")))) = "0.008061 moles I"_2

2.558 color(red)(cancel(color(black)("g"))) * "1 mole ZnI"_2/(319.22color(red)(cancel(color(black)("g")))) = "0.008013 moles ZnI"_2

Notice that you have more moles of zinc metal than moles of iodine, which means that iodine will act as a limiting reagent, i.e. it will be completely consumed before all the moles of zinc get a chance to react.

The $1 : 1$ mole ratio that exists between the two tells you that the reaction will consume $0.008061$ moles of iodine and

0.008061 color(red)(cancel(color(black)("moles I"_2))) * "1 mole Zn"/(1color(red)(cancel(color(black)("mole I"_2)))) = "0.008061 moles Zn"

Now, the reaction can theoretically produce $1$ mole of zinc iodide for every $1$ mole of zinc and $1$ mole of iodine consumed.

You can thus say that theoretically, the reaction should produce

0.008061color(red)(cancel(color(black)("moles I"_2))) * "1 mole ZnI"_2/(1color(red)(cancel(color(black)("mole I"_2)))) = "0.008061 moles ZnI"_2

However, you know that the reaction actually produced $0.008013$ moles of zinc iodide.

This of course implies that the reaction does not have a 100% yield, i.e. not all the moles of zinc and iodine that react are converted to moles of zinc iodide.

The percent yield of the reaction is given by

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "what you actually get"/"what you could theoretically get} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, this amounts to

"% yield" = (0.008013 color(red)(cancel(color(black)("moles ZnI"_2))))/(0.008061color(red)(cancel(color(black)("moles ZnI"_2)))) xx 100 = color(green)(bar(ul(|color(white)(a/a)color(black)(99.4%)color(white)(a/a)|)))#

The answer is rounded to three sig figs.