Question

Question #c429b

Answer 1

`"% yield" = 99.4%`

Explanation:

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

`"Zn"_ ((s)) + "I"_ (2(aq)) -> "ZnI"_ (2(aq))`

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

`color(blue)(bar(ul(|color(white)(a/a)"% yield" = "what you actually get"/"what you could theoretically get" xx 100 color(white)(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.