Question #88ad7
1 Answer
Explanation:
Water decomposes according to the following chemical equation
#color(blue)(2)"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))#
Notice that it takes
#(color(blue)(2)color(white)(.)"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)#
This means that for every
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
Now, you know that the produced
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.