Question #6547e

1 Answer
Jun 11, 2017

Answer:

I get 111.25%.

Explanation:

What is oleum?

Oleum is a solution of #"SO"_3# in #"H"_2"SO"_4#.

Addition of water to oleum converts the free #"SO"_3# into #"H"_2"SO"_4#, and the resulting solution will contain only #"H"_2"SO"_4#.

#"SO"_3 + "H"_2"O" → "H"_2"SO"_4#

What is your sample?

Your sample contains 50% "free #"SO"_3#" and 50 % "combined #"SO"_3"#" as #"H"_2"SO"_4#.

How do we express #"SO"_3# as % oleum?

% Oleum is the mass of water required to react with the free #"SO"_3# in 100 g of oleum.

The reaction is

#M_text(r): 80.06color(white)(m)18.02color(white)(mml) 98.08#
#color(white)(mmll)"SO"_3 + "H"_2"O" → "H"_2"SO"_4#

If we use 100 g of your oleum, it will contain 50 g of #"SO"_3#

#"Mass of H"_2"O" = 50 color(red)(cancel(color(black)("g SO"_3))) × ("18.02 g H"_2"O")/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "11.25 g H"_2"O"#

So, the sample would be classed as 11.25 % oleum or 111.25 % #"H"_2"SO"_4#.