Equal weights of phosphorus and oxygen are heated in a closed vessel producing `P_(2)O_(3) and P_(2)O_(5)` in a 1 : 1 mole ratio. If the limiting component is exhausted, find which component and also what fraction of it is left over.

Explanation

As 1 mole each of `P_(2) O_(3) and P_(2) O_(5)` is formed, the mole ratio of the reacting P and O atoms is 4 : 8 or 1 : 2 As the given weights of phosphorus and oxygen are same, say w g, their
Moles of atoms are `(omega)/(31) and (omega)/(16)` respectively.
`:.` 2 moles of O combine with 1 mole of P.
`:. (omega)/(16)` moles of O combine with `(omega)/( 32)` moles of P.
So phosphorous is left over and
moles of P remaining `= (omega)/( 31) - (omega)/( 32)`
`:.` fraction of P left over `=(((omega)/(31)-(omega)/(32)))/((omega)/(31))= (1)/(32) or 3 . 125 %`