Question #e31e4
1 Answer
Explanation:
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
#2"Pb"_ ((s)) + "O"_ (2(g)) -> 2"PbO"_ ((s))#
Notice that you have a
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
#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.