Question #c06a7
1 Answer
Here's what's going on here.
Explanation:
Oleum is simply a saturated solution of sulfur trioxide,
You'll sometimes see this oleum referred to as disulfuric acid,
Oleum will always have a percentage of sulfuric acid that is greater than
#"SO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_2"SO"_text(4(aq])#
In essence, the concentration of oleum tells you how much water must combine with the free sulfur trioxide present in
Let's take, for example, a
Take a look at the above equation. Notice that one mole of sulfur trioxide will react with one mole of water to form one mole of sulfuric acid.
Use the molar masses of the three compounds to determine the ratio that exists between their masses.
#"For SO"_3: " "" "M_M ~~ "80 g mol"^(-1)#
#"For H"_2"O": " "" "M_M ~~ "18 g mol"^(-1)#
#"For H"_2"SO"_4:" " M_M ~~ "98 g mol"^(-1)#
So, this tells you that
In this example,
#5 color(red)(cancel(color(black)("g water"))) * "80 g SO"_3/(18color(red)(cancel(color(black)("g water")))) = "22.22 g SO"_3#
This means that
The percent of free sulfur trioxide will thus be
Implicitly the percent of sulfuric acid in this oleum will be
#"% H"_2"SO"_4 = 100 - 22.22 = 77.78%#
Notice that this oleum solution will produce
#5color(red)(cancel(color(black)("g water"))) * ("98 g H"_2"SO"_4)/(18color(red)(cancel(color(black)("g water")))) = "27.22 g H"_2"SO"_4#
Since the
#"77.78 g " + " 27.22 g" = "105 g H"_2"SO"_4#
So, as a conclusion, the percentage of free sulfur trioxide tells you much sulfur trioxide you get per