Question #697de

1 Answer
Oct 10, 2016

Answer:

#62%#

Explanation:

The first thing to do here is to figure out the theoretical yield of the reaction, which is essentially what you get for a reaction that has an #100%# yield.

#"C"_ 2"H"_ (4(g)) + "H"_ 2"O"_ ((g)) -> "CH"_ 3"CH"_ 2"OH"_ ((g))#

Notice that every mole of ethene that takes part in the reaction produces #1# mole of ethanol. In your case, you have

#4.8 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_4)/(28.05color(red)(cancel(color(black)("g")))) = "0.1711 moles C"_2"H"_4#

The #1:1# mole ratio that exists between ethene and ethanol tells you that the reaction can theoretically produce #0.1711# moles of ethenol, the equivalent of

#0.1711 color(red)(cancel(color(black)("moles CH"_3"CH"_2"OH"))) * "46.07 g"/(1color(red)(cancel(color(black)("mole CH"_3"CH"_2"OH")))) = "7.88 g"#

As you can see, the actual yield of the reaction is lower than the theoretical yield. To find the percent yield simply use

#color(purple)(bar(ul(|color(white)(a/a)color(black)("% yield" = "actual yield"/"theoretical yield" xx 100%)color(white)(a/a)|)))#

Plug in your values to find

#"% yield" = (4.9 color(red)(cancel(color(black)("g"))))/(7.88color(red)(cancel(color(black)("g")))) xx 100% = color(green)(bar(ul(|color(white)(a/a)color(black)(62%)color(white)(a/a)|)))#

The answer is rounded to two sig figs.