Oleum is mixture of `H_(2)SO_(4)` and `SO_(3)` i.e. `H_(2)S_(2)O_(7)` which is obtained by passing `SO_(3)` is solution of `H_(2)SO_(4)`. In order to dissolve `SO_(3)` in oleum, dilution of oleum is done by water in which oleum is converted into pure `H_(2)SO` as shown below:
`H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4)` (pure)
When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum.
For example: `109% H_(2)SO_(4)` labelling of oleum sample means that 109 gm pure `H_(2)SO_(4)` is obtained on diluting 100 gm oleum with 9 gm `H_(2)O` which dissolves al free `SO_(3)` in oleum.
If the number of moles of free `SO_(3), H_(2)SO_(4)`, and `H_(2)O` be x, y and z respectively in 118%` H_(2)SO_(4)` labelled oleum, the value of `(x+y+z)` is

Options

A. 2.2
B. 3.2
C. 3.4
D. 4.2

Answer

A