Question #c429b
1 Answer
Explanation:
Start by writing the balanced chemical equation that describes this synthesis reaction
#"Zn"_ ((s)) + "I"_ (2(aq)) -> "ZnI"_ (2(aq))#
You know that
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
#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
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
This of course implies that the reaction does not have a
The percent yield of the reaction is given by
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.