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100 gms of 118% oleum reacted with exces...

100 gms of `118%` oleum reacted with excess water. How much `H_(2)SO_(4)` is formed?

A

90 gm

B

118 gm

C

48 gm

D

100 gm

Text Solution

Verified by Experts

The correct Answer is:
B
`SO_(3)+H_(2)SO_(4) to H_(2)S_(2)O_(7), H_(2)S_(2)O_(7)+H_(2)O to 2H_(2)SO_(4)`
`SO_(3)` is absorbed in `98%` conc. `H_(2)SO_(4)` to get oleum. Oleum `[H_(2)S_(2)O_(7)]` is diluted with water to get `H_(2)SO_(4)`
118% of oleum `to` 100 gm
`" ? " larr 100%`, 118 grams.
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Knowledge Check

  • Oleum or fuming H_(2)SO_(4) is

    A
    A mixture of conc. `H_(2)SO_(4)` and oil
    B
    Sulphuric acid which gives fumes of sulphur dioxide
    C
    Sulphuric acid saturated with sulphur trioxide, i.e., `H_(2)S_(2)O_(7)`
    D
    A mixture of sulphuric acid and nitric acid

  • Cl_2 reacts with water and forms

    A
    `HCl + HOCl`
    B
    `HCl + O_2 + O_3`
    C
    `HCl + HOCl + O_3`
    D
    `HOCl + O_2`

  • Bleaching powder when treated with excess of dil. H_(2) SO_(4) liberates

    A
    `Cl_(2)`
    B
    `O_(2)`
    C
    both `O_(2)` and `Cl_(2)`
    D
    None

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    Which of the following react with dilute H_(2)SO_(4) to form H_(2) .

    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

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