Question #c429b

1 Answer
Sep 22, 2016

Answer:

#"% 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.