Question

`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`?

Answer 1

`67.5%`

Explanation:

You know by looking at the balanced chemical equation

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

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

`("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)`

So, you know that every `"36.03 g"` of water that take part in the reaction produce `"32.0 g"` of oxygen gas.

This means that when `"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 `"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.