Calculate the moles of NaOH required to neutralize the solution produced by dissolving 1.1 g `P_(4)O_(6)` in water. Use the following reactions:
`P_(4)O_(6) + 6H_(2)O to 4H_(3)PO_(3), 2NaOH + H_(3)PO_(3) to Na_(2)HPO_(3) + 2H_(2)O` (At. Mass/g `mol^(-1)`P=31, O=16).

Explanation

`P_(4)O_(6) + 6H_(2)O to 4H_(3)PO_(3), 2NaOH + H_(3)PO_(3) to Na_(2)HPO_(3) + 2H_(2)O`
Molecular weight of `P_(4)O_(6) = 4 xx 31 + 6 xx 16 = 124 + 96 = 220 g mol^(-1)`
Number of moles of `P_(4)O_(6) = ("Mass given")/("Molecular Mass") = (1.1g)/220 = 1/200` moles
`1/200` moles of `P_(4)O_(6)` produces `4 xx 1/200` moles of `H_(3)PO_(4)`
1 mole of `H_(3)PO_(4)` requires 2 moles of NaOH.
`1/50` moles of `H_(3)PO_(4)` requires `2 xx 1/50 = 1/25 = 0.04` moles of NaOH.