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Similar to % labelling of oleum ,a mixt...

Similar to % labelling of oleum ,a mixture of `H_(3)PO_(3)` and `P_(4)O_(6)` is labelled as (100 +x)% where x is maximum mass of water which can reacts with `P_(4)O_(6)` present in 100 gm mixture of `H_(3)PO_(3)` and `P_(4)O_(6)`
`P_(4)O_(6) + H_(2)O rarr H_(3)PO_(3)`
If such a mixture is labelled as 127 % , then mass of free `P_(4)O_(6)` in
given 100 g mixture is :

A

45 gm

B

55 gm

C

71 gm

D

58 gm

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Similar to % labelling of oleum ,a mixture of H_(3)PO_(3) and P_(4)O_(6) is labelled as (100 +x)% where x is maximum mass of water which can reacts with P_(4)O_(6) present in 100 gm mixture of H_(3)PO_(3) and P_(4)O_(6) P_(4)O_(6) + H_(2)O rarr H_(3)PO_(3) For such a mixture of P_(4)O_(6) and H_(3)PO_(3) labelled as (100 +x)% . Value of x can lie in range of (maximum and minimum) :

underline(P_(4))O_(6)+6H _(2)O to 4H_(3)PO_(3)

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H_(4)underline(P_(2))O_()+H_(2)O to H_(3)PO_(3)

underline(P_(4))O_(10)+H_(2)O to H_(3)PO_(4)

H_(4)underline(P_(2))O_(7)+H_(2)O to 2H_(3)PO_(4)

H_(4)underline(P_(2))O_(7)+H_(2)O to 2H_(3)PO_(4)

An oleum sample is labelled as 118%, Calculate Mass of H_(2)SO_(4) in 100 gm oleum sample

H_(4)underline(P_(2))O_(6)+H_(2)O to H_(3)PO_(3)+H_(3)PO_(4)

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