Question #e31e4

1 Answer
Oct 16, 2016


#"% yield" = 73.05%#


The first thing to do here is to calculate the theoretical yield of the reaction, i.e. the amount of lead(II) oxide formed for a reaction that has a #100%# yield.

#2"Pb"_ ((s)) + "O"_ (2(g)) -> 2"PbO"_ ((s))#

Notice that you have a #2:2# mole ratio between lead and lead(II) oxide, which means that the reaction will theoretically produce as many moles of lead(II) oxide as you have moles of lead taking part in the reaction.

Use the molar mass of lead to convert the mass to moles

#451.4 color(red)(cancel(color(black)("g"))) * "1 mole Pb"/(207.2color(red)(cancel(color(black)("g")))) = "2.1786 moles Pb"#

You now know that the theoretical yield of the reaction is equal to

#2.1786 color(red)(cancel(color(black)("moles Pb"))) * "2 moles PbO"/(2color(red)(cancel(color(black)("moles Pb")))) = "2.1786 moles PbO"#

which is equivalent to

#2.1786 color(red)(cancel(color(black)("moles PbO"))) * "223.2 g"/(1color(red)(cancel(color(black)("mole PbO")))) = "486.26 g"#

Now, the actual yield of the reaction is #"355.2 g"# of lead(II) oxide. The percent yield is defined as

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

In your case, the percent yield of the reaction will be

#color(darkgreen)(bar(ul(|color(white)(a/a)color(black)("% yield" = (color(blue)(355.2 color(red)(cancel(color(blue)("g")))))/(color(purple)(486.26color(red)(cancel(color(purple)("g"))))) xx 100% = 73.05%)color(white)(a/a)|)))#

The answer is rounded to four sig figs.