# Question 697de

Oct 10, 2016

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.

${\text{C"_ 2"H"_ (4(g)) + "H"_ 2"O"_ ((g)) -> "CH"_ 3"CH"_ 2"OH}}_{\left(g\right)}$

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.