#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
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 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
This means that when
#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
Now, you know that the actual yield of the reaction is
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.