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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)`
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) :

A

`0 lt x lt 22.5`

B

`0 lt x lt 49.09`

C

`0 lt x lt 59.65`

D

`0 lt x lt 34.66`

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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) If such a mixture is labelled as 127 % , then mass of free P_(4)O_(6) in given 100 g mixture is :

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

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

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)

For H_(3)PO_(3) and H_(3)PO_(4) the correct choice is

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

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