# Question 88ad7

Mar 11, 2017

66.2%

#### Explanation:

Water decomposes according to the following chemical equation

$\textcolor{b l u e}{2} {\text{H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O}}_{2 \left(g\right)}$

Notice that it takes $\textcolor{b l u e}{2}$ 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

$\left(\textcolor{b l u e}{2} \textcolor{w h i t e}{.} {\text{moles H"_2"O")/"1 mole O"_2 = (color(blue)(2) color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O}}_{2}\right)$

This means that for every $\text{36.03 g}$ of water that undergo decomposition, the reaction can theoretically produce $\text{32.0 g}$ 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 $\text{20.4 g}$ 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.