#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#?

1 Answer
Jun 9, 2017

Answer:

#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.