Question

Question #88ad7

Answer 1

`66.2%`

Explanation:

Water decomposes according to the following chemical equation

`color(blue)(2)"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))`

Notice that it takes `color(blue)(2)` moles of water to produce `1` mole of oxygen gas. You can convert this mole ratio to a gram ratio by using the molar masses of the two chemical species

`(color(blue)(2)color(white)(.)"moles H"_2"O")/"1 mole O"_2 = (color(blue)(2) color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O"_2)`

This means that for every `"36.03 g"` of water that undergo decomposition, the reaction can theoretically produce `"32.0 g"` of oxygen gas.

In your case, the sample of water that underwent decomposition should have produced

`34.7 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "30.8 g O"_2`

This represents the reaction's theoretical yield, i.e. what you expect to get for a `100%` yield.

Now, you know that the produced `"20.4 g"` of oxygen gas. This represents the reaction's actual yield, i.e. what you actually get by performing the reaction.

The reaction's percent yield can be calculated by using the equation

`color(blue)(ul(color(black)("% yield" = "actual yield"/"theoretical yield" xx 100%)))`

Plug in your values to find

`"% yield" = (20.4 color(red)(cancel(color(black)("g"))))/(30.8color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(66.2%)))`

The answer is rounded to three sig figs.