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) :

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 :

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 :

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) (100)/(3) gm mixture of H_(3)PO_(3) and P_(4)O_(6) (labelled as 123%) is mixed with 50 L of H_(2)O and resulting solution (assuming volume change) is mixed with 2 M NaOH solution . Volume of NaOH required for complete neutralisationn is:

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

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

P_(4)+NaOH+H_(2)O rarr NaH_(2)PO_(3)+PH_(3) is

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)

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