# 2H_2O -> 2H_2 + O_2 What is the percent yield of O_2 if 10.2 g of O_2 is produced from the decomposition of 17.0 g of H_2O?

Jun 9, 2017

67.5%

#### Explanation:

You know by looking at the balanced chemical equation

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

that it takes $2$ moles of water to produce $1$ mole of oxygen gas. Use the molar masses of the two chemical species to convert this mole ratio to a gram ratio

$\left({\text{2 moles H"_2"O")/("1 mole O"_2) = (2 color(red)(cancel(color(black)("moles H"_2"O"))) * ("18.015 g H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g O"_ 2/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O}}_{2}\right)$

So, you know that every $\text{36.03 g}$ of water that take part in the reaction produce $\text{32.0 g}$ of oxygen gas.

This means that when $\text{17.0 g}$ of water undergo decomposition, you should expect to get

17.0 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "15.1 g O"_2

This represents the reaction's theoretical yield, i.e. what is produced by a reaction that has a 100% yield.

Now, you know that the actual yield of the reaction is $\text{10.2 g}$ of oxygen gas. To find the percent yield, divide the actual yield by the theoretical yield and multiply the result by 100%.

You should get

"% yield" = (10.2 color(red)(cancel(color(black)("g O"_2))))/(15.1color(red)(cancel(color(black)("g O"_2)))) * 100% = color(darkgreen)(ul(color(black)(67.5%)))

The answer is rounded to three sig figs.